Okay, so I need to find the smallest positive perfect cube that can be written as the sum of three consecutive integers. Let me start by understanding the problem. A perfect cube is a number like 1, 8, 27, 64, etc., where it's an integer multiplied by itself three times. Three consecutive integers would be numbers like n, n+1, n+2 for some integer n. So the sum would be n + (n+1) + (n+2) = 3n + 3. That simplifies to 3(n + 1). So the sum is three times some integer (n + 1). 

So the problem reduces to finding the smallest perfect cube that is a multiple of 3. Because 3(n + 1) is a multiple of 3, so the perfect cube must be divisible by 3. Let's list the perfect cubes and see which is the smallest that's divisible by 3.

The perfect cubes in order are 1, 8, 27, 64, 125, 216, etc. Now, check which of these are divisible by 3. 1 divided by 3 is not an integer. 8 divided by 3 is not an integer. 27 divided by 3 is 9, which is an integer. So 27 is the first perfect cube that is divisible by 3. Therefore, the sum 3(n + 1) = 27. Then n + 1 = 9, so n = 8. Therefore, the three consecutive integers are 8, 9, 10. Their sum is 27, which is indeed a perfect cube. Wait, but let me check if there's a perfect cube smaller than 27 that can be expressed as such a sum. The next smaller cube is 8. If 3(n + 1) = 8, then n + 1 = 8/3 ≈ 2.666, which is not an integer, so n would not be an integer. Similarly, 1 is too small. So 27 is the next candidate. 

But wait, let me verify. Suppose the three consecutive integers are n-1, n, n+1. Then their sum is 3n. So if the sum is a perfect cube, then 3n = k^3. Then n = k^3 / 3. For n to be an integer, k^3 must be divisible by 3, so k must be a multiple of 3. Let k = 3m. Then k^3 = 27m^3. So 3n = 27m^3, so n = 9m^3. Therefore, the three integers would be 9m^3 - 1, 9m^3, 9m^3 + 1. Their sum is 27m^3. The smallest positive m is 1, so the sum is 27*1 = 27. Therefore, the smallest perfect cube is 27. 

But wait, let me check another approach. Maybe the three consecutive integers don't have to be centered around zero. Wait, but the problem didn't specify that they have to be positive, but since the sum is a positive perfect cube, the integers themselves could be negative or positive. But since we want the smallest positive perfect cube, which is 1, 8, 27, etc., and the sum is 3n+3, which is 3(n+1). So as long as 3(n+1) is a perfect cube, n can be negative. For example, if n+1 is 1, then n=0, sum is 3. But 3 isn't a perfect cube. If n+1 is 8, then n=7, sum is 24, which isn't a cube. Wait, maybe I confused something here. 

Wait, if the sum is 3(n+1) = k^3, then k^3 must be a multiple of 3. So the smallest k^3 that is a multiple of 3 is 27, since 3^3=27. Therefore, the sum is 27, and the integers are n, n+1, n+2 where 3n + 3 = 27, so 3n = 24, n=8. So numbers are 8,9,10. Sum is 27. 

Alternatively, if I take three consecutive numbers starting from a negative integer. For example, -1, 0, 1. Their sum is 0, which is 0^3, but we need positive. Next, 0,1,2 sum to 3. Not a cube. 1,2,3 sum to 6. Not a cube. 2,3,4 sum to 9. Not a cube. 3,4,5 sum to 12. Not a cube. 4,5,6 sum to 15. Not a cube. 5,6,7 sum to 18. Not a cube. 6,7,8 sum to 21. Not a cube. 7,8,9 sum to 24. Not a cube. 8,9,10 sum to 27. That's the first one. So 27 is indeed the answer. 

But let me check if there's a smaller cube. The next cube after 1 is 8. Let's see if 8 can be expressed as the sum of three consecutive integers. Let's set up the equation: n + (n+1) + (n+2) = 3n + 3 = 8. Then 3n = 5, so n = 5/3. Not an integer. So 8 is not possible. Then 27 is possible. 

Therefore, the answer should be 27. But wait, let me check if there's another way. Suppose the three consecutive integers are not starting from n, but maybe something else. Wait, no, any three consecutive integers can be written as n-1, n, n+1, so their sum is 3n. So 3n must equal k^3. Therefore, k^3 must be a multiple of 3, so k must be a multiple of 3. So the smallest k is 3, so k^3 is 27. Therefore, the smallest positive perfect cube is 27. 

Therefore, the answer is 27. But let me check once more. Let me compute the sum of 8,9,10: 8+9=17, 17+10=27. Yes, that's correct. And 27 is 3^3, which is a perfect cube. 

Is there any smaller positive perfect cube that can be expressed as such a sum? Let's check 1: 1. The sum of three consecutive integers would have to be 1. Let's see: n + (n+1) + (n+2) = 3n + 3 = 1. Then 3n = -2, n = -2/3. Not integer. So no. Next, 8. As before, n would be 5/3. Not integer. So next is 27. So 27 is indeed the smallest. 

Therefore, the answer is 27. 

Wait, but let me think again. The problem says "three consecutive integers". So could they be negative? For example, if we take three consecutive negative integers, their sum could be positive. Let's see: for example, -1, 0, 1. Sum is 0. But 0 isn't positive. Then 0,1,2: sum 3. Not a cube. 1,2,3: sum 6. 2,3,4: sum 9. 3,4,5: sum 12. 4,5,6: sum 15. 5,6,7: sum 18. 6,7,8: sum 21. 7,8,9: sum 24. 8,9,10: sum 27. So yes, 27 is the first one. So even if we consider negative integers, the sum can't be 1 or 8, the next is 27. 

Alternatively, if we take the numbers as 0,1,2: sum 3. 1,2,3: sum 6. 2,3,4: sum 9. 3,4,5: sum 12. 4,5,6: sum 15. 5,6,7: sum 18. 6,7,8: sum 21. 7,8,9: sum 24. 8,9,10: sum 27. So same result. 

Therefore, I think 27 is the correct answer. 

But let me verify once more with algebra. Let the three consecutive integers be x, x+1, x+2. Their sum is 3x + 3. We need 3x + 3 = k^3, where k is a positive integer. Then 3(x + 1) = k^3. Therefore, k^3 must be divisible by 3. Therefore, k must be divisible by 3, since 3 is prime. Let k = 3m. Then k^3 = 27m^3. So 3(x + 1) = 27m^3. Therefore, x + 1 = 9m^3, so x = 9m^3 - 1. Therefore, the three integers are 9m^3 - 1, 9m^3, 9m^3 + 1. So the sum is 27m^3. The smallest positive perfect cube is when m=1: 27*1=27. Therefore, the integers are 8,9,10. Sum is 27. 

Yes, this confirms the answer is 27. 

I don't see any mistakes in this reasoning. So the answer should be 27.
Therefore, the smallest positive perfect cube that can be written as the sum of three consecutive integers is $\boxed{27}$.
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