💡Okay, so I need to find the expected value of rolling an eight-sided die. Hmm, I remember that expected value is like the average outcome you'd expect if you rolled the die many times. But I'm not entirely sure how to calculate it exactly. Let me think.First, the die has eight faces, numbered from 1 to 8. So, the possible outcomes are 1, 2, 3, 4, 5, 6, 7, and 8. Since it's a fair die, each of these outcomes has an equal chance of happening, right? That means each number has a probability of 1/8.Now, expected value is calculated by multiplying each outcome by its probability and then adding all those products together. So, in formula terms, it should be something like:E = (1 * P(1)) + (2 * P(2)) + ... + (8 * P(8))Since each probability P(n) is 1/8, this simplifies to:E = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) * (1/8)Okay, so I need to add up all the numbers from 1 to 8 first. Let me do that step by step to make sure I don't make a mistake.1 + 2 = 33 + 3 = 66 + 4 = 1010 + 5 = 1515 + 6 = 2121 + 7 = 2828 + 8 = 36Wait, that doesn't seem right. Let me check again.1 + 2 = 33 + 3 = 66 + 4 = 1010 + 5 = 1515 + 6 = 2121 + 7 = 2828 + 8 = 36Hmm, I get 36 both times. Maybe that's correct. I think there's a formula for the sum of the first n natural numbers, which is n(n + 1)/2. Let's apply that to verify.n = 8, so the sum should be 8 * 9 / 2 = 72 / 2 = 36. Okay, that matches. So the sum of numbers from 1 to 8 is indeed 36.Now, going back to the expected value formula:E = 36 * (1/8) = 36/8Let me compute that. 36 divided by 8 is 4.5.So, the expected value is 4.5. That makes sense because it's the average of the numbers from 1 to 8. Since 4.5 is halfway between 4 and 5, it feels right.I think I'm confident with this answer now. The expected value of rolling an eight-sided die is 4.5.

💡First, I recognize that the two pots and their shadows form similar triangles because the angle of elevation of the sun is the same for both pots.Given that the shorter pot is 20 inches tall and casts a 10-inch shadow, I can set up a proportion to find the shadow length of the taller pot, which is 40 inches tall.By setting up the proportion ( frac{20}{10} = frac{40}{S_2} ), I can solve for ( S_2 ), the shadow length of the taller pot.Simplifying the proportion gives me ( 2 = frac{40}{S_2} ), and solving for ( S_2 ) results in ( S_2 = 20 ) inches.Therefore, the taller pot casts a shadow that is 20 inches long.

💡Alright, so I'm trying to figure out this question about the emergence of "Digital Earth" and what it means in terms of technological innovation. The options are A, B, C, and D, and I need to pick the correct one. Let me break it down step by step.First, I need to understand what "Digital Earth" is. From what I remember, Digital Earth is a concept that involves creating a comprehensive digital representation of Earth's physical and human systems. It uses technologies like GIS (Geographic Information Systems), remote sensing, and the internet to integrate and visualize data about the planet. So, it's a pretty advanced and integrated technological initiative.Now, the question says that the emergence of Digital Earth is "another paradigm of technological innovation." Paradigms are like frameworks or models that define how we think about and approach things. So, if Digital Earth is a new paradigm, it must represent a significant shift in how we innovate technologically.Looking at the options:A: Technological innovation is increasingly globalized.B: Science and technology are the concentrated expression of advanced productive forces.C: The development of science and technology cannot be separated from strong economic strength.D: All technology serves politics.I need to see which of these aligns with the idea that Digital Earth is a new paradigm in technological innovation.Starting with option A: "Technological innovation is increasingly globalized." This makes sense because Digital Earth involves integrating data from various parts of the world, using technologies developed by different countries and organizations. It's a collaborative effort on a global scale. So, this seems plausible.Option B: "Science and technology are the concentrated expression of advanced productive forces." This sounds a bit abstract. Productive forces refer to the means and methods of producing goods and services. Science and technology do drive productivity, but I'm not sure if this directly relates to Digital Earth being a new paradigm in innovation.Option C: "The development of science and technology cannot be separated from strong economic strength." While economic strength does play a role in funding technological projects, I'm not sure if this is the main point about Digital Earth being a new paradigm.Option D: "All technology serves politics." This seems too broad and not directly related to the concept of Digital Earth as a technological innovation paradigm.Going back to option A, the idea of globalization in technological innovation fits well with Digital Earth. It's not just a local or national project; it's something that requires global cooperation and data sharing. This aligns with the notion of a new paradigm because it represents a shift towards more integrated, global approaches to technology development.But wait, option B mentions "advanced productive forces." Maybe that's also relevant because Digital Earth enhances our ability to manage and utilize resources more effectively, which ties into productive forces. However, I think the key here is the paradigm shift towards globalization in innovation.So, after considering all options, I think option A is the best fit because it directly addresses the global nature of technological innovation exemplified by Digital Earth.

💡Okay, so I have this quadratic equation: 4a x squared minus 2a x plus c equals zero, and it's given that one of the roots is zero. I need to find the other root. Hmm, let's see. First, I remember that for a quadratic equation of the form ax² + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a. Maybe I can use that here. Wait, but in this equation, the coefficients are a bit different. The equation is 4a x² - 2a x + c = 0. So, comparing it to the standard form, the coefficient of x² is 4a, the coefficient of x is -2a, and the constant term is c. Since one of the roots is zero, let's denote the roots as 0 and t, where t is the other root we need to find. Using the sum of the roots formula: 0 + t = -(-2a)/(4a). Let me compute that. The negative of -2a is 2a, and then divided by 4a. So, 2a divided by 4a simplifies to 2/4, which is 1/2. So, t equals 1/2. Wait, that seems straightforward. But let me double-check using the product of the roots. The product should be 0 * t = c/(4a). But 0 times t is 0, so c/(4a) should be 0. That means c must be zero. Hold on, if c is zero, then the original equation becomes 4a x² - 2a x = 0. We can factor out 2a x, giving 2a x (2x - 1) = 0. So, the roots are x = 0 and x = 1/2. That matches what I found earlier. Hmm, so both methods give me the same result. The other root is 1/2. I think that's correct. But just to be thorough, let me plug x = 1/2 back into the original equation to see if it satisfies it. Plugging in x = 1/2: 4a*(1/2)^2 - 2a*(1/2) + c. Let's compute each term. First term: 4a*(1/4) = a. Second term: -2a*(1/2) = -a. Third term: c. So, adding them up: a - a + c = 0 + c. But earlier, we found that c must be zero. So, 0 + 0 = 0, which satisfies the equation. Okay, that checks out. So, the other root is definitely 1/2. I think I'm confident with this answer now. It makes sense both algebraically and when plugging it back into the equation. Plus, using both the sum and product of roots confirmed the same result. So, yeah, the other root is 1/2.

💡Okay, so I have this problem where I need to figure out how many different ways I can express the number 121 as the sum of distinct Fibonacci numbers. The Fibonacci sequence starts with 1, 1, and each subsequent number is the sum of the two before it. So, the sequence goes 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, and so on. First, I need to list out all the Fibonacci numbers that are less than or equal to 121. Let me do that:1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.Wait, I see that there are two 1s at the beginning. But since we're talking about distinct Fibonacci numbers, I guess we can only use each number once. So, I should probably consider the Fibonacci sequence without repeating the 1. So, maybe it's just 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.Now, I need to find all the possible ways to add up these numbers to get 121. Since the order doesn't matter, it's more about combinations rather than permutations. I remember something called Zeckendorf's theorem, which states that every positive integer can be uniquely represented as the sum of one or more distinct non-consecutive Fibonacci numbers. But I'm not sure if that helps here because the problem doesn't specify that the Fibonacci numbers need to be non-consecutive. It just says they have to be distinct. So, maybe there are multiple ways to represent 121 as the sum of distinct Fibonacci numbers.Let me start by trying to use the largest Fibonacci number less than or equal to 121, which is 89. If I subtract 89 from 121, I get 32. Now, I need to express 32 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 13, 21, 34, 55.Wait, 34 is larger than 32, so I can't use that. The next largest is 21. If I subtract 21 from 32, I get 11. Now, I need to express 11 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 13.Again, 13 is too big, so the next is 8. Subtracting 8 from 11 gives me 3. Now, I need to express 3 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5.3 is in the list, so I can use that. So, one way to express 121 is 89 + 21 + 8 + 3.Let me write that down: 89 + 21 + 8 + 3 = 121.Is there another way to express 32? Let's see. Instead of 21 + 8 + 3, maybe I can use smaller Fibonacci numbers.Starting again with 32. The largest Fibonacci number less than or equal to 32 is 21. If I use 21, then I have 11 left. As before, 11 can be expressed as 8 + 3, but I already did that. Alternatively, can I express 11 differently?11 can also be 5 + 5 + 1, but wait, we can't use the same Fibonacci number more than once. So, 5 + 5 is not allowed. What about 8 + 2 + 1? That adds up to 11. So, another way to express 32 is 21 + 8 + 2 + 1.So, another way to express 121 is 89 + 21 + 8 + 2 + 1 = 121.Okay, that's a second way.Is there another way to express 32? Let's see. Instead of 21, maybe I can use 13. So, 32 - 13 = 19. Now, I need to express 19 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 21.The largest Fibonacci number less than or equal to 19 is 13, but I already used 13. Wait, no, I haven't used 13 yet in this path. Wait, no, in this case, I used 13 first, so I can't use it again. So, the next largest is 8. 19 - 8 = 11. Now, I need to express 11 as before.11 can be 8 + 2 + 1, but I already used 8 in this path, so I can't use it again. Alternatively, 5 + 3 + 2 + 1 = 11. So, another way is 13 + 8 + 5 + 3 + 2 + 1 = 32.Wait, but 13 + 8 + 5 + 3 + 2 + 1 = 32. So, another way to express 121 is 89 + 13 + 8 + 5 + 3 + 2 + 1 = 121.That's a third way.Is there another way? Let's see. Maybe instead of 13 + 8 + 5 + 3 + 2 + 1, can I use different combinations?Alternatively, 32 can be expressed as 21 + 8 + 3, which we already did, or 21 + 5 + 3 + 2 + 1, which is another way. Wait, 21 + 5 + 3 + 2 + 1 = 32. So, that's another way to express 32.So, another way to express 121 is 89 + 21 + 5 + 3 + 2 + 1 = 121.That's a fourth way.Is there another way? Let's see. Maybe using 13 instead of 21.32 - 13 = 19. Then, 19 can be expressed as 8 + 5 + 3 + 2 + 1. So, that's similar to what I did before.Alternatively, 32 can be expressed as 13 + 5 + 3 + 2 + 1 + 8, but that's the same as before.Wait, maybe I can use 55 instead of 89. Let's try that.121 - 55 = 66. Now, I need to express 66 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 13, 21, 34, 89.But 89 is already used, so I can't use it again. The largest Fibonacci number less than or equal to 66 is 55, but I already used 55. So, the next is 34. 66 - 34 = 32. Now, I need to express 32 as before.We already have ways to express 32, so 55 + 34 + 21 + 8 + 3 = 121.Wait, 55 + 34 = 89, and then 89 + 21 + 8 + 3 = 121. But that's the same as the first way I did it, just using 55 and 34 instead of 89.Wait, no, because 55 + 34 is 89, so it's the same as just using 89. So, that doesn't give a new way.Alternatively, 55 + 34 + 21 + 8 + 3 is the same as 89 + 21 + 8 + 3, so it's not a new way.Hmm, maybe using 55 and then 34 and then smaller numbers.55 + 34 = 89, and then 89 + 21 + 8 + 3 = 121, which is the same as before.Alternatively, 55 + 21 = 76, and then 121 - 76 = 45. Now, I need to express 45 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 13, 34, 89.The largest Fibonacci number less than or equal to 45 is 34. 45 - 34 = 11. Now, 11 can be expressed as 8 + 2 + 1. So, another way is 55 + 21 + 34 + 8 + 2 + 1 = 121.Wait, that's a different combination. So, that's a fifth way.Is that correct? Let me check: 55 + 21 + 34 + 8 + 2 + 1 = 55 + 21 = 76, 76 + 34 = 110, 110 + 8 = 118, 118 + 2 = 120, 120 + 1 = 121. Yes, that adds up.So, that's a fifth way.Is there another way using 55? Let's see. 55 + 13 = 68, and then 121 - 68 = 53. Now, 53 can be expressed as 34 + 13 + 5 + 1, but I already used 13. Wait, no, I haven't used 13 yet in this path. Wait, I used 13 in the initial step, so I can't use it again. So, 53 can be expressed as 34 + 8 + 5 + 3 + 2 + 1. So, another way is 55 + 13 + 34 + 8 + 5 + 3 + 2 + 1 = 121.Wait, but 55 + 13 + 34 + 8 + 5 + 3 + 2 + 1 = 55 + 13 = 68, 68 + 34 = 102, 102 + 8 = 110, 110 + 5 = 115, 115 + 3 = 118, 118 + 2 = 120, 120 + 1 = 121. That works.So, that's a sixth way.Is there another way? Let's see. Maybe using 55 + 8 = 63, and then 121 - 63 = 58. Now, 58 can be expressed as 34 + 21 + 3. So, another way is 55 + 8 + 34 + 21 + 3 = 121.Let me check: 55 + 8 = 63, 63 + 34 = 97, 97 + 21 = 118, 118 + 3 = 121. Yes, that works.So, that's a seventh way.Is there another way? Let's see. Maybe using 55 + 5 = 60, and then 121 - 60 = 61. 61 can be expressed as 34 + 21 + 5 + 1, but I already used 5. So, 61 can be 34 + 21 + 5 + 1, but since 5 is already used, I can't do that. Alternatively, 61 can be 34 + 13 + 8 + 5 + 1, but I haven't used 13 or 8 yet. So, 55 + 5 + 34 + 13 + 8 + 1 = 121.Wait, let's check: 55 + 5 = 60, 60 + 34 = 94, 94 + 13 = 107, 107 + 8 = 115, 115 + 1 = 116. Wait, that's not 121. Hmm, maybe I made a mistake.Wait, 55 + 5 + 34 + 13 + 8 + 1 = 55 + 5 = 60, 60 + 34 = 94, 94 + 13 = 107, 107 + 8 = 115, 115 + 1 = 116. That's only 116, not 121. So, that doesn't work. Maybe I need to adjust.Alternatively, 61 can be expressed as 34 + 21 + 5 + 1, but since I used 5, I can't. Maybe 34 + 13 + 8 + 5 + 1, but again, 5 is used. Hmm, maybe this path doesn't work.Alternatively, maybe using 55 + 3 = 58, and then 121 - 58 = 63. 63 can be expressed as 34 + 21 + 8. So, another way is 55 + 3 + 34 + 21 + 8 = 121.Let me check: 55 + 3 = 58, 58 + 34 = 92, 92 + 21 = 113, 113 + 8 = 121. Yes, that works.So, that's an eighth way.Is there another way? Let's see. Maybe using 55 + 2 = 57, and then 121 - 57 = 64. 64 can be expressed as 34 + 21 + 8 + 1. So, another way is 55 + 2 + 34 + 21 + 8 + 1 = 121.Wait, let's check: 55 + 2 = 57, 57 + 34 = 91, 91 + 21 = 112, 112 + 8 = 120, 120 + 1 = 121. Yes, that works.So, that's a ninth way.Is there another way? Let's see. Maybe using 55 + 1 = 56, and then 121 - 56 = 65. 65 can be expressed as 34 + 21 + 8 + 2. So, another way is 55 + 1 + 34 + 21 + 8 + 2 = 121.Checking: 55 + 1 = 56, 56 + 34 = 90, 90 + 21 = 111, 111 + 8 = 119, 119 + 2 = 121. Yes, that works.So, that's a tenth way.Is there another way? Let's see. Maybe using 34 instead of 55. So, 121 - 34 = 87. Now, I need to express 87 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 13, 21, 55, 89.The largest Fibonacci number less than or equal to 87 is 55. 87 - 55 = 32. Now, I need to express 32 as before.We already have ways to express 32, so 34 + 55 + 21 + 8 + 3 = 121.Wait, that's similar to the first way, just starting with 34 instead of 89. So, it's not a new way.Alternatively, 34 + 55 + 13 + 8 + 5 + 3 + 2 + 1 = 121, which is similar to the sixth way.Hmm, maybe I'm not getting any new ways here.Let me try another approach. Maybe using 21 as the largest Fibonacci number.121 - 21 = 100. Now, I need to express 100 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 13, 34, 55, 89.The largest Fibonacci number less than or equal to 100 is 89. 100 - 89 = 11. Now, 11 can be expressed as 8 + 2 + 1. So, another way is 21 + 89 + 8 + 2 + 1 = 121.Wait, that's similar to the second way I did earlier, just starting with 21 instead of 89. So, it's not a new way.Alternatively, 100 can be expressed as 55 + 34 + 8 + 3. So, another way is 21 + 55 + 34 + 8 + 3 = 121.Checking: 21 + 55 = 76, 76 + 34 = 110, 110 + 8 = 118, 118 + 3 = 121. Yes, that works.But that's similar to the fifth way I did earlier, just starting with 21 instead of 55. So, it's not a new way.Hmm, I'm not finding any new ways here.Let me try using 13 as the largest Fibonacci number.121 - 13 = 108. Now, I need to express 108 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 8, 21, 34, 55, 89.The largest Fibonacci number less than or equal to 108 is 89. 108 - 89 = 19. Now, 19 can be expressed as 13 + 5 + 1, but I already used 13. So, 19 can be 8 + 5 + 3 + 2 + 1. So, another way is 13 + 89 + 8 + 5 + 3 + 2 + 1 = 121.But that's similar to the sixth way I did earlier, just starting with 13 instead of 55. So, it's not a new way.Alternatively, 108 can be expressed as 55 + 34 + 13 + 8 + 5 + 3 + 2 + 1, but that's the same as before.Hmm, I'm not finding any new ways here either.Let me try using 8 as the largest Fibonacci number.121 - 8 = 113. Now, I need to express 113 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 5, 13, 21, 34, 55, 89.The largest Fibonacci number less than or equal to 113 is 89. 113 - 89 = 24. Now, 24 can be expressed as 21 + 3. So, another way is 8 + 89 + 21 + 3 = 121.But that's similar to the first way I did earlier, just starting with 8 instead of 89. So, it's not a new way.Alternatively, 24 can be expressed as 13 + 8 + 3, but I already used 8. So, 24 can be 13 + 5 + 3 + 2 + 1. So, another way is 8 + 89 + 13 + 5 + 3 + 2 + 1 = 121.But that's similar to the sixth way I did earlier, just starting with 8 instead of 55. So, it's not a new way.Hmm, I'm not finding any new ways here either.Let me try using 5 as the largest Fibonacci number.121 - 5 = 116. Now, I need to express 116 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 3, 8, 13, 21, 34, 55, 89.The largest Fibonacci number less than or equal to 116 is 89. 116 - 89 = 27. Now, 27 can be expressed as 21 + 5 + 1, but I already used 5. So, 27 can be 13 + 8 + 5 + 1, but I already used 5. Alternatively, 27 can be 21 + 3 + 2 + 1. So, another way is 5 + 89 + 21 + 3 + 2 + 1 = 121.But that's similar to the second way I did earlier, just starting with 5 instead of 89. So, it's not a new way.Alternatively, 27 can be expressed as 13 + 8 + 5 + 1, but since I used 5, I can't. So, maybe 13 + 8 + 3 + 2 + 1 = 27. So, another way is 5 + 89 + 13 + 8 + 3 + 2 + 1 = 121.But that's similar to the sixth way I did earlier, just starting with 5 instead of 55. So, it's not a new way.Hmm, I'm not finding any new ways here either.Let me try using 3 as the largest Fibonacci number.121 - 3 = 118. Now, I need to express 118 as the sum of distinct Fibonacci numbers from the remaining list: 1, 2, 5, 8, 13, 21, 34, 55, 89.The largest Fibonacci number less than or equal to 118 is 89. 118 - 89 = 29. Now, 29 can be expressed as 21 + 8. So, another way is 3 + 89 + 21 + 8 = 121.But that's similar to the first way I did earlier, just starting with 3 instead of 89. So, it's not a new way.Alternatively, 29 can be expressed as 13 + 8 + 5 + 3, but I already used 3. So, 29 can be 13 + 8 + 5 + 3, but since I used 3, I can't. Alternatively, 29 can be 13 + 8 + 5 + 3, but again, 3 is used. Hmm, maybe this path doesn't work.Alternatively, 29 can be expressed as 21 + 5 + 3, but I already used 3. So, maybe this path doesn't work.Hmm, I'm not finding any new ways here either.Let me try using 2 as the largest Fibonacci number.121 - 2 = 119. Now, I need to express 119 as the sum of distinct Fibonacci numbers from the remaining list: 1, 3, 5, 8, 13, 21, 34, 55, 89.The largest Fibonacci number less than or equal to 119 is 89. 119 - 89 = 30. Now, 30 can be expressed as 21 + 5 + 3 + 1. So, another way is 2 + 89 + 21 + 5 + 3 + 1 = 121.But that's similar to the second way I did earlier, just starting with 2 instead of 89. So, it's not a new way.Alternatively, 30 can be expressed as 13 + 8 + 5 + 3 + 1. So, another way is 2 + 89 + 13 + 8 + 5 + 3 + 1 = 121.But that's similar to the sixth way I did earlier, just starting with 2 instead of 55. So, it's not a new way.Hmm, I'm not finding any new ways here either.Finally, let me try using 1 as the largest Fibonacci number.121 - 1 = 120. Now, I need to express 120 as the sum of distinct Fibonacci numbers from the remaining list: 2, 3, 5, 8, 13, 21, 34, 55, 89.The largest Fibonacci number less than or equal to 120 is 89. 120 - 89 = 31. Now, 31 can be expressed as 21 + 8 + 2. So, another way is 1 + 89 + 21 + 8 + 2 = 121.But that's similar to the second way I did earlier, just starting with 1 instead of 89. So, it's not a new way.Alternatively, 31 can be expressed as 13 + 8 + 5 + 3 + 2. So, another way is 1 + 89 + 13 + 8 + 5 + 3 + 2 = 121.But that's similar to the sixth way I did earlier, just starting with 1 instead of 55. So, it's not a new way.Hmm, I think I've exhausted all possibilities. Let me count how many ways I found:1. 89 + 21 + 8 + 32. 89 + 21 + 8 + 2 + 13. 89 + 13 + 8 + 5 + 3 + 2 + 14. 89 + 21 + 5 + 3 + 2 + 15. 55 + 34 + 21 + 8 + 36. 55 + 13 + 34 + 8 + 5 + 3 + 2 + 17. 55 + 8 + 34 + 21 + 38. 55 + 3 + 34 + 21 + 89. 55 + 2 + 34 + 21 + 8 + 110. 55 + 1 + 34 + 21 + 8 + 2Wait, that's ten ways. But earlier, I thought it was eight. Maybe I made a mistake in counting.Let me list them again:1. 89 + 21 + 8 + 32. 89 + 21 + 8 + 2 + 13. 89 + 13 + 8 + 5 + 3 + 2 + 14. 89 + 21 + 5 + 3 + 2 + 15. 55 + 34 + 21 + 8 + 36. 55 + 13 + 34 + 8 + 5 + 3 + 2 + 17. 55 + 8 + 34 + 21 + 38. 55 + 3 + 34 + 21 + 89. 55 + 2 + 34 + 21 + 8 + 110. 55 + 1 + 34 + 21 + 8 + 2Yes, that's ten ways. But I'm not sure if all of them are valid because some might have overlapping numbers or might not be distinct.Wait, let me check each one:1. 89 + 21 + 8 + 3 = 121. All distinct.2. 89 + 21 + 8 + 2 + 1 = 121. All distinct.3. 89 + 13 + 8 + 5 + 3 + 2 + 1 = 121. All distinct.4. 89 + 21 + 5 + 3 + 2 + 1 = 121. All distinct.5. 55 + 34 + 21 + 8 + 3 = 121. All distinct.6. 55 + 13 + 34 + 8 + 5 + 3 + 2 + 1 = 121. All distinct.7. 55 + 8 + 34 + 21 + 3 = 121. All distinct.8. 55 + 3 + 34 + 21 + 8 = 121. All distinct.9. 55 + 2 + 34 + 21 + 8 + 1 = 121. All distinct.10. 55 + 1 + 34 + 21 + 8 + 2 = 121. All distinct.Yes, all of them are valid and use distinct Fibonacci numbers. So, I think the answer is 10 ways.Wait, but I'm not sure if I'm missing any ways or if some of these are duplicates. Let me see:- Ways 1 and 5 are similar but use different combinations.- Ways 2 and 4 are similar but use different combinations.- Ways 3 and 6 are similar but use different combinations.- Ways 7, 8, 9, and 10 are all different.I think they are all unique. So, the total number of ways is 10.But wait, I'm not sure if the problem considers the order of addition. The problem says, "The same addends in a different order are considered the same way." So, as long as the set of numbers is the same, regardless of order, it's considered one way. So, in my list above, each way is a unique set of numbers, so they are all distinct.Therefore, the answer should be 10 ways.But wait, I'm not sure if I'm overcounting. Let me see:- 89 + 21 + 8 + 3- 89 + 21 + 8 + 2 + 1- 89 + 13 + 8 + 5 + 3 + 2 + 1- 89 + 21 + 5 + 3 + 2 + 1- 55 + 34 + 21 + 8 + 3- 55 + 13 + 34 + 8 + 5 + 3 + 2 + 1- 55 + 8 + 34 + 21 + 3- 55 + 3 + 34 + 21 + 8- 55 + 2 + 34 + 21 + 8 + 1- 55 + 1 + 34 + 21 + 8 + 2Yes, each of these is a unique combination of Fibonacci numbers adding up to 121. So, I think the answer is 10.But I'm not entirely sure because sometimes in these problems, certain combinations might not be valid or might overlap. Maybe I should double-check.Let me check each combination:1. 89 + 21 + 8 + 3 = 121. Correct.2. 89 + 21 + 8 + 2 + 1 = 121. Correct.3. 89 + 13 + 8 + 5 + 3 + 2 + 1 = 121. Correct.4. 89 + 21 + 5 + 3 + 2 + 1 = 121. Correct.5. 55 + 34 + 21 + 8 + 3 = 121. Correct.6. 55 + 13 + 34 + 8 + 5 + 3 + 2 + 1 = 121. Correct.7. 55 + 8 + 34 + 21 + 3 = 121. Correct.8. 55 + 3 + 34 + 21 + 8 = 121. Correct.9. 55 + 2 + 34 + 21 + 8 + 1 = 121. Correct.10. 55 + 1 + 34 + 21 + 8 + 2 = 121. Correct.Yes, all of them add up correctly. So, I think the answer is 10 ways.But wait, I'm not sure if the problem allows using the same Fibonacci number more than once. The problem says "sum of distinct Fibonacci numbers," so I think each number can only be used once. In my combinations above, I made sure to use each Fibonacci number only once. So, that's correct.Therefore, I think the answer is 10 different ways.

💡Alright, so I have this problem where I need to find a third-degree equation with integer coefficients, and the roots of this equation are sin(π/14), sin(5π/14), and sin(-3π/14). Hmm, okay, let's break this down step by step.First, I remember that for a polynomial with roots r1, r2, r3, the polynomial can be written as (x - r1)(x - r2)(x - r3). So, if I can find expressions for these sine terms and then expand the product, I should get the desired polynomial. But since the coefficients need to be integers, I might need to manipulate these sine expressions into a form that allows me to do that.Looking at the angles, π/14, 5π/14, and -3π/14. I notice that these angles are all related to π/14, which is 12.857 degrees. So, π/14 is a relatively small angle, and 5π/14 is about 64.285 degrees, while -3π/14 is negative, but sine is an odd function, so sin(-3π/14) is equal to -sin(3π/14). So, maybe I can rewrite the roots as sin(π/14), sin(5π/14), and -sin(3π/14).Hmm, so the roots are sin(π/14), sin(5π/14), and -sin(3π/14). I wonder if there's a way to relate these sine terms to each other or to some known trigonometric identities that can help me find a polynomial equation.I recall that there are multiple-angle formulas for sine and cosine, and sometimes these can be used to relate different sine terms. Maybe I can use the identity for sin(7θ) since 7 times π/14 is π/2, which is a nice angle. Let me write that down:sin(7θ) = 7sinθ - 56sin³θ + 112sin⁵θ - 64sin⁷θ.Wait, is that right? Let me check. I think the expansion for sin(7θ) is indeed a polynomial in sinθ. Let me verify the coefficients. I think it's:sin(7θ) = 7sinθ - 56sin³θ + 112sin⁵θ - 64sin⁷θ.Yes, that seems correct. So, if I set θ = π/14, then 7θ = π/2, and sin(π/2) = 1. So, plugging θ = π/14 into the equation:1 = 7sin(π/14) - 56sin³(π/14) + 112sin⁵(π/14) - 64sin⁷(π/14).Hmm, so that gives me an equation involving sin(π/14). Similarly, if I set θ = 5π/14, then 7θ = 5π/2, and sin(5π/2) = 1 as well. So, plugging θ = 5π/14:1 = 7sin(5π/14) - 56sin³(5π/14) + 112sin⁵(5π/14) - 64sin⁷(5π/14).And for θ = 3π/14, 7θ = 3π/2, and sin(3π/2) = -1. So, plugging θ = 3π/14:-1 = 7sin(3π/14) - 56sin³(3π/14) + 112sin⁵(3π/14) - 64sin⁷(3π/14).But since one of our roots is -sin(3π/14), let's denote x = sin(3π/14), so -x = sin(-3π/14). Then, the equation becomes:-1 = 7x - 56x³ + 112x⁵ - 64x⁷.Multiplying both sides by -1:1 = -7x + 56x³ - 112x⁵ + 64x⁷.Hmm, that's a bit messy, but maybe I can relate these equations together. Let me think.So, I have three equations:1. For sin(π/14):1 = 7sin(π/14) - 56sin³(π/14) + 112sin⁵(π/14) - 64sin⁷(π/14).2. For sin(5π/14):1 = 7sin(5π/14) - 56sin³(5π/14) + 112sin⁵(5π/14) - 64sin⁷(5π/14).3. For sin(3π/14):-1 = 7sin(3π/14) - 56sin³(3π/14) + 112sin⁵(3π/14) - 64sin⁷(3π/14).But since we have roots at sin(π/14), sin(5π/14), and -sin(3π/14), which is equivalent to -sin(3π/14). So, maybe I can consider the minimal polynomial for sin(π/14), sin(5π/14), and sin(3π/14), and then adjust for the negative sign.Alternatively, perhaps there is a way to express these sine terms in terms of cosines, since sometimes cosine polynomials are easier to handle. I remember that sin(θ) = cos(π/2 - θ). So, let's try that.So, sin(π/14) = cos(π/2 - π/14) = cos(3π/7).Similarly, sin(5π/14) = cos(π/2 - 5π/14) = cos(π/7).And sin(3π/14) = cos(π/2 - 3π/14) = cos(2π/7).Therefore, our roots are cos(3π/7), cos(π/7), and -cos(2π/7).Wait, but we have -sin(3π/14) = -cos(2π/7). So, the roots are cos(3π/7), cos(π/7), and -cos(2π/7).Hmm, okay, so maybe I can find a polynomial whose roots are cos(π/7), cos(3π/7), and cos(5π/7), and then adjust for the negative sign.I recall that there is a known polynomial for cos(π/7), cos(3π/7), and cos(5π/7). Let me try to recall it.I think it's 8x³ - 4x² - 4x + 1 = 0. Wait, is that correct? Let me check.Alternatively, I think the minimal polynomial for cos(2π/7) is 8x³ + 4x² - 4x - 1 = 0. Hmm, but I'm not sure. Maybe I need to derive it.Alternatively, I can use the identity for cos(7θ) = 0, since 7θ = π, 3π, 5π, etc., but that might complicate things.Wait, let's think differently. Since cos(7θ) = 0 when θ = π/14, 3π/14, 5π/14, etc. So, maybe I can use the expansion for cos(7θ) and set it equal to zero.The expansion for cos(7θ) is:cos(7θ) = 64cos⁷θ - 112cos⁵θ + 56cos³θ - 7cosθ.So, setting cos(7θ) = 0, we get:64cos⁷θ - 112cos⁵θ + 56cos³θ - 7cosθ = 0.Factor out cosθ:cosθ(64cos⁶θ - 112cos⁴θ + 56cos²θ - 7) = 0.So, the solutions are cosθ = 0, or 64cos⁶θ - 112cos⁴θ + 56cos²θ - 7 = 0.But we are interested in θ = π/14, 3π/14, 5π/14, etc., which are not multiples of π/2, so cosθ ≠ 0. Therefore, the non-zero solutions satisfy:64cos⁶θ - 112cos⁴θ + 56cos²θ - 7 = 0.Let me denote y = cos²θ. Then, the equation becomes:64y³ - 112y² + 56y - 7 = 0.So, 64y³ - 112y² + 56y - 7 = 0.Hmm, that's a cubic in y. Maybe I can factor this or find rational roots.Let me try rational root theorem. Possible rational roots are factors of 7 over factors of 64, so ±1, ±7, ±1/2, ±7/2, etc. Let's test y = 1:64(1) - 112(1) + 56(1) - 7 = 64 - 112 + 56 - 7 = 1. Not zero.y = 7: That's too big, probably not.y = 1/2:64*(1/8) - 112*(1/4) + 56*(1/2) - 7 = 8 - 28 + 28 - 7 = 1. Not zero.y = 7/2: That's 3.5, which is too big.y = 1/4:64*(1/64) - 112*(1/16) + 56*(1/4) - 7 = 1 - 7 + 14 - 7 = 1. Not zero.Hmm, not promising. Maybe it's irreducible. So, perhaps the minimal polynomial for cos²(π/14) is this cubic.But I'm not sure if that helps me directly. Maybe I need another approach.Wait, earlier I thought about expressing the sines as cosines. So, our roots are cos(3π/7), cos(π/7), and -cos(2π/7). So, if I can find a polynomial with these roots, that would be the minimal polynomial.I recall that the minimal polynomial for cos(2π/7) is 8x³ + 4x² - 4x - 1 = 0. Let me check that.If I plug x = cos(2π/7) into 8x³ + 4x² - 4x - 1, does it equal zero?I don't remember exactly, but let's assume for a moment that it is. Then, similarly, cos(π/7) and cos(3π/7) would be roots of the same polynomial, right? Because they are all related through the heptagon angles.Wait, actually, cos(π/7), cos(3π/7), and cos(5π/7) are the roots of the same minimal polynomial. Let me confirm.Yes, I think the minimal polynomial for cos(2π/7) is indeed 8x³ + 4x² - 4x - 1 = 0. So, if that's the case, then the roots are cos(2π/7), cos(4π/7), and cos(6π/7). Wait, but 4π/7 is greater than π/2, so cos(4π/7) is negative, and similarly for cos(6π/7). Hmm, but in our case, our roots are cos(π/7), cos(3π/7), and -cos(2π/7). So, maybe they are related.Wait, cos(π/7) is approximately 0.9009688679, cos(3π/7) is approximately 0.2225209339, and cos(2π/7) is approximately 0.623489802. So, they are all positive, but our third root is -cos(2π/7), which is negative.So, perhaps the minimal polynomial for cos(π/7), cos(3π/7), and cos(5π/7) is different. Wait, cos(5π/7) is equal to -cos(2π/7), since cos(π - x) = -cos(x). So, cos(5π/7) = -cos(2π/7). Therefore, the roots are cos(π/7), cos(3π/7), and -cos(2π/7).So, if I can find a polynomial with these roots, that would be the minimal polynomial. Let me denote the roots as a = cos(π/7), b = cos(3π/7), c = -cos(2π/7).So, the polynomial would be (x - a)(x - b)(x - c). Let's expand this.First, multiply (x - a)(x - b):= x² - (a + b)x + ab.Then, multiply by (x - c):= (x² - (a + b)x + ab)(x - c)= x³ - (a + b + c)x² + (ab + ac + bc)x - abc.So, the coefficients are:- Coefficient of x²: -(a + b + c)- Coefficient of x: ab + ac + bc- Constant term: -abcSo, if I can find the values of a + b + c, ab + ac + bc, and abc, I can write the polynomial.But calculating these directly might be complicated. Maybe there's a trigonometric identity that can help.I recall that for angles that are multiples of π/7, there are certain sum identities. Let me see.First, let's compute a + b + c = cos(π/7) + cos(3π/7) - cos(2π/7).Hmm, is there a known sum for these cosines?I think there is a formula for the sum of cosines of angles in arithmetic progression. The formula is:cos A + cos(A + D) + cos(A + 2D) + ... + cos(A + (n-1)D) = [sin(nD/2) / sin(D/2)] * cos(A + (n-1)D/2).But in our case, we have three terms: cos(π/7), cos(3π/7), and -cos(2π/7). It's not a straightforward arithmetic progression, but maybe we can manipulate it.Alternatively, perhaps we can use the identity for sum of cosines:cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2).But with three terms, it's more complicated. Maybe I can pair two terms first.Let me try:cos(π/7) + cos(3π/7) = 2 cos((π/7 + 3π/7)/2) cos((π/7 - 3π/7)/2) = 2 cos(2π/7) cos(-π/7) = 2 cos(2π/7) cos(π/7).Since cosine is even, cos(-x) = cos(x).So, cos(π/7) + cos(3π/7) = 2 cos(2π/7) cos(π/7).Therefore, a + b + c = 2 cos(2π/7) cos(π/7) - cos(2π/7) = cos(2π/7)(2 cos(π/7) - 1).Hmm, not sure if that helps directly, but maybe we can find the value of this expression.Alternatively, perhaps I can use known values or identities for these specific angles.Wait, I remember that cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2. Is that correct?Let me check:Using the identity for sum of cosines:cos(π/7) + cos(3π/7) + cos(5π/7) = ?I think it's indeed 1/2. Let me verify.Yes, I recall that the sum of cos(π/7), cos(3π/7), and cos(5π/7) is 1/2. So, that means:cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2.But in our case, c = -cos(2π/7), and cos(5π/7) = -cos(2π/7). So, cos(5π/7) = c.Therefore, cos(π/7) + cos(3π/7) + c = 1/2.So, a + b + c = 1/2.Okay, that's useful. So, the sum of the roots is 1/2.Now, moving on to the next coefficient: ab + ac + bc.So, ab + ac + bc = cos(π/7)cos(3π/7) + cos(π/7)(-cos(2π/7)) + cos(3π/7)(-cos(2π/7)).Simplify:= cos(π/7)cos(3π/7) - cos(π/7)cos(2π/7) - cos(3π/7)cos(2π/7).Hmm, this seems complicated, but maybe we can use product-to-sum identities.Recall that cos A cos B = [cos(A+B) + cos(A-B)] / 2.So, let's compute each term:1. cos(π/7)cos(3π/7) = [cos(4π/7) + cos(-2π/7)] / 2 = [cos(4π/7) + cos(2π/7)] / 2.2. cos(π/7)cos(2π/7) = [cos(3π/7) + cos(-π/7)] / 2 = [cos(3π/7) + cos(π/7)] / 2.3. cos(3π/7)cos(2π/7) = [cos(5π/7) + cos(π/7)] / 2.So, substituting back:ab + ac + bc = [cos(4π/7) + cos(2π/7)] / 2 - [cos(3π/7) + cos(π/7)] / 2 - [cos(5π/7) + cos(π/7)] / 2.Let's combine these terms:= [cos(4π/7) + cos(2π/7) - cos(3π/7) - cos(π/7) - cos(5π/7) - cos(π/7)] / 2.Simplify the numerator:cos(4π/7) + cos(2π/7) - cos(3π/7) - 2cos(π/7) - cos(5π/7).Hmm, let's see if we can simplify this further.First, note that cos(4π/7) = cos(π - 3π/7) = -cos(3π/7).Similarly, cos(5π/7) = cos(π - 2π/7) = -cos(2π/7).So, substituting these:= [-cos(3π/7) + cos(2π/7) - cos(3π/7) - 2cos(π/7) - (-cos(2π/7))] / 2.Simplify:= [-cos(3π/7) + cos(2π/7) - cos(3π/7) - 2cos(π/7) + cos(2π/7)] / 2.Combine like terms:= [-2cos(3π/7) + 2cos(2π/7) - 2cos(π/7)] / 2.Factor out the 2:= 2[-cos(3π/7) + cos(2π/7) - cos(π/7)] / 2.Cancel the 2:= -cos(3π/7) + cos(2π/7) - cos(π/7).Hmm, this seems to be going in circles. Maybe there's another approach.Alternatively, perhaps I can use the identity that the sum of the products of the roots two at a time is equal to (sum of roots)^2 - sum of squares of roots, all divided by 2.Wait, no, that's for symmetric sums. Let me recall:For a cubic equation x³ + px² + qx + r = 0, the sum of roots is -p, the sum of products two at a time is q, and the product of roots is -r.But in our case, the polynomial is (x - a)(x - b)(x - c) = x³ - (a + b + c)x² + (ab + ac + bc)x - abc.So, we have:- Coefficient of x²: -(a + b + c) = -1/2.- Coefficient of x: ab + ac + bc.- Constant term: -abc.So, if I can find ab + ac + bc and abc, I can write the polynomial.Alternatively, maybe I can find these values using known identities or by using the minimal polynomial.Wait, earlier I thought that the minimal polynomial for cos(2π/7) is 8x³ + 4x² - 4x - 1 = 0. If that's the case, then the roots are cos(2π/7), cos(4π/7), and cos(6π/7). But in our case, the roots are cos(π/7), cos(3π/7), and -cos(2π/7). So, maybe there's a relationship between these polynomials.Alternatively, perhaps I can use the fact that cos(π/7), cos(3π/7), and cos(5π/7) are roots of a certain polynomial, and then adjust for the negative sign.Wait, I think the minimal polynomial for cos(2π/7) is indeed 8x³ + 4x² - 4x - 1 = 0. So, if I can relate our roots to this polynomial, maybe I can find the desired polynomial.But our roots are cos(π/7), cos(3π/7), and -cos(2π/7). So, if I let x = cos(π/7), y = cos(3π/7), z = -cos(2π/7).Then, the polynomial would be (x - a)(x - b)(x - c) where a, b, c are these roots.But I need to find the coefficients in terms of a, b, c.Alternatively, maybe I can use the fact that cos(π/7), cos(3π/7), and cos(5π/7) are roots of the polynomial 8x³ - 4x² - 4x + 1 = 0. Wait, is that correct?Let me check. If I plug x = cos(π/7) into 8x³ - 4x² - 4x + 1, does it equal zero?I don't remember exactly, but let's assume for a moment that it is. Then, the polynomial would be 8x³ - 4x² - 4x + 1 = 0.But in our case, one of the roots is -cos(2π/7), which is equal to cos(5π/7), since cos(5π/7) = -cos(2π/7). So, if cos(5π/7) is a root of 8x³ - 4x² - 4x + 1 = 0, then our roots are cos(π/7), cos(3π/7), and cos(5π/7), which are all roots of this polynomial.Wait, but in our problem, the roots are sin(π/14), sin(5π/14), and sin(-3π/14), which we converted to cos(3π/7), cos(π/7), and -cos(2π/7). So, cos(3π/7) is equal to sin(π/14), cos(π/7) is equal to sin(5π/14), and -cos(2π/7) is equal to sin(-3π/14).Therefore, the minimal polynomial for these roots is the same as the minimal polynomial for cos(π/7), cos(3π/7), and cos(5π/7), which is 8x³ - 4x² - 4x + 1 = 0.Wait, but earlier I thought the minimal polynomial for cos(2π/7) is 8x³ + 4x² - 4x - 1 = 0. So, perhaps there's a sign difference because of the negative root.Let me double-check.If I consider the polynomial 8x³ - 4x² - 4x + 1 = 0, and plug in x = cos(π/7), does it satisfy?I think it does, but I'm not 100% sure. Alternatively, perhaps I can derive it.Let me try to derive the minimal polynomial for cos(π/7).We know that cos(7θ) = 0 when θ = π/14, 3π/14, 5π/14, etc. So, using the expansion for cos(7θ):cos(7θ) = 64cos⁷θ - 112cos⁵θ + 56cos³θ - 7cosθ = 0.So, for θ = π/14, we have:64cos⁷(π/14) - 112cos⁵(π/14) + 56cos³(π/14) - 7cos(π/14) = 0.But cos(π/14) = sin(π/2 - π/14) = sin(6π/14) = sin(3π/7). Wait, that's not helpful.Alternatively, perhaps I can express cos(π/7) in terms of cos(π/14). Hmm, not sure.Alternatively, let me consider that cos(π/7) is related to the roots of the equation 8x³ - 4x² - 4x + 1 = 0.Let me assume that 8x³ - 4x² - 4x + 1 = 0 is the minimal polynomial for cos(π/7), cos(3π/7), and cos(5π/7). Then, since our roots are exactly these cosines, the polynomial would be 8x³ - 4x² - 4x + 1 = 0.But wait, in our case, one of the roots is negative: -cos(2π/7) = cos(5π/7). So, if cos(5π/7) is a root of 8x³ - 4x² - 4x + 1 = 0, then the polynomial remains the same.Therefore, the polynomial with roots sin(π/14), sin(5π/14), and sin(-3π/14) is 8x³ - 4x² - 4x + 1 = 0.Wait, but earlier I thought the minimal polynomial for cos(2π/7) is 8x³ + 4x² - 4x - 1 = 0. So, there might be a sign difference because of the negative root.Let me check both polynomials.If I plug x = cos(π/7) into 8x³ - 4x² - 4x + 1, does it equal zero?I don't have the exact value, but let's approximate:cos(π/7) ≈ 0.9009688679.Compute 8*(0.9009688679)^3 - 4*(0.9009688679)^2 - 4*(0.9009688679) + 1.First, 0.9009688679^3 ≈ 0.730.8*0.730 ≈ 5.84.0.9009688679^2 ≈ 0.811.4*0.811 ≈ 3.244.4*0.9009688679 ≈ 3.604.So, 5.84 - 3.244 - 3.604 + 1 ≈ 5.84 - 3.244 = 2.596; 2.596 - 3.604 = -1.008; -1.008 + 1 = -0.008.Hmm, that's close to zero, but not exactly. Maybe due to approximation errors.Alternatively, let's try x = cos(3π/7) ≈ 0.2225209339.Compute 8*(0.2225209339)^3 - 4*(0.2225209339)^2 - 4*(0.2225209339) + 1.0.2225209339^3 ≈ 0.011.8*0.011 ≈ 0.088.0.2225209339^2 ≈ 0.0495.4*0.0495 ≈ 0.198.4*0.2225209339 ≈ 0.890.So, 0.088 - 0.198 - 0.890 + 1 ≈ 0.088 - 0.198 = -0.11; -0.11 - 0.890 = -1.00; -1.00 + 1 = 0.Wow, that's exactly zero. So, x = cos(3π/7) is a root of 8x³ - 4x² - 4x + 1 = 0.Similarly, let's try x = cos(5π/7) ≈ -0.623489803.Compute 8*(-0.623489803)^3 - 4*(-0.623489803)^2 - 4*(-0.623489803) + 1.First, (-0.623489803)^3 ≈ -0.242.8*(-0.242) ≈ -1.936.(-0.623489803)^2 ≈ 0.388.4*0.388 ≈ 1.552.4*(-0.623489803) ≈ -2.493959212.So, -1.936 - 1.552 + 2.493959212 + 1 ≈ (-1.936 - 1.552) = -3.488; (-3.488 + 2.494) ≈ -0.994; (-0.994 + 1) ≈ 0.006.Again, close to zero, likely due to approximation errors.Therefore, it seems that 8x³ - 4x² - 4x + 1 = 0 is indeed the minimal polynomial for cos(π/7), cos(3π/7), and cos(5π/7).Since our roots are exactly these cosines, the polynomial with roots sin(π/14), sin(5π/14), and sin(-3π/14) is 8x³ - 4x² - 4x + 1 = 0.Wait, but earlier I thought the minimal polynomial for cos(2π/7) is 8x³ + 4x² - 4x - 1 = 0. So, why is there a difference?I think it's because the minimal polynomial depends on the specific roots. In our case, the roots are cos(π/7), cos(3π/7), and cos(5π/7), which are all distinct and correspond to different angles. The polynomial 8x³ - 4x² - 4x + 1 = 0 is the minimal polynomial for these roots.Therefore, the desired polynomial with integer coefficients having roots sin(π/14), sin(5π/14), and sin(-3π/14) is 8x³ - 4x² - 4x + 1 = 0.But wait, let me double-check the signs. Our roots are sin(π/14) = cos(3π/7), sin(5π/14) = cos(π/7), and sin(-3π/14) = -sin(3π/14) = -cos(2π/7) = cos(5π/7). So, yes, the roots are cos(3π/7), cos(π/7), and cos(5π/7), which are the roots of 8x³ - 4x² - 4x + 1 = 0.Therefore, the polynomial is 8x³ - 4x² - 4x + 1 = 0.But wait, in the earlier step, when I plugged in x = cos(5π/7) ≈ -0.623489803, the result was approximately 0.006, which is very close to zero, considering the approximation errors. So, it seems consistent.Therefore, the final answer is 8x³ - 4x² - 4x + 1 = 0.