<Question>
What is the smallest positive perfect cube that can be written as the sum of three consecutive integers?
</Question>

<think>
<Plan_1>
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 step by step.

<prompt_1.1> Understand the problem and set up the equation for the sum of three consecutive integers. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, a perfect cube is a number like 1, 8, 27, 64, 125, etc. , which are cubes of integers 1, 2, 3, 4, 5, and so on. The question is asking for the smallest such cube that can be expressed as the sum of three consecutive integers. 

Let me denote the three consecutive integers. If I let the first integer be \( n \), then the next two would be \( n + 1 \) and \( n + 2 \). So, their sum would be \( n + (n + 1) + (n + 2) \). Let me compute that: that's \( 3n + 3 \), which simplifies to \( 3(n + 1) \). Wait, so the sum of three consecutive integers is always three times the middle number, which makes sense because the middle number is the average of the three.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, the sum is \( 3(n + 1) \). Therefore, we need this sum to be a perfect cube. So, \( 3(n + 1) = k^3 \), where \( k \) is some positive integer. Then, solving for \( n \), we get \( n = \frac{k^3}{3} - 1 \).

<prompt_2.1> Derive the expression for \( n \) in terms of \( k \) and consider the conditions for \( n \) to be an integer. </prompt_2.1>

Let me check with the smallest perfect cubes:

<prompt_2.2> Check the smallest perfect cubes to find the smallest one that can be expressed as the sum of three consecutive integers. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Since \( n \) has to be an integer (because we're dealing with integers), \( \frac{k^3}{3} \) must also be an integer. Therefore, \( k^3 \) must be divisible by 3. Since 3 is a prime number, for \( k^3 \) to be divisible by 3, \( k \) itself must be divisible by 3. (Because if a prime divides a cube, it must divide the base). So, \( k \) must be a multiple of 3. Let me write \( k \) as \( 3m \), where \( m \) is a positive integer. Then, substituting back, we have:

\( 3(n + 1) = (3m)^3 = 27m^3 \)

Divide both sides by 3:

\( n + 1 = 9m^3 \)

Therefore, \( n = 9m^3 - 1 \)

So, the three consecutive integers would be \( 9m^3 - 1 \), \( 9m^3 \), and \( 9m^3 + 1 \). Their sum is \( 27m^3 \), which is \( (3m)^3 \). Therefore, the sum is a perfect cube. 

But wait, we need the smallest positive perfect cube. So, the smallest \( m \) is 1. Let's check that.
</execution_2.1>

<execution_2.2>
Starting with 1: If \( k = 1 \), then \( k^3 = 1 \). Then \( n = \frac{1}{3} - 1 = -\frac{2}{3} \). But \( n \) needs to be an integer, so this doesn't work.

Next, \( k = 2 \): \( 2^3 = 8 \). Then \( n = \frac{8}{3} - 1 = \frac{5}{3} \). Still not an integer. Hmm.

Next, \( k = 3 \): \( 3^3 = 27 \). Then \( n = \frac{27}{3} - 1 = 9 - 1 = 8 \). That's an integer! So, if \( n = 8 \), then the three consecutive integers are 8, 9, 10. Their sum is 27, which is indeed \( 3^3 \).
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.2:
But wait, is 27 the smallest possible?

Wait, let's check if there is a smaller cube than 27 that can be expressed as the sum of three consecutive integers. The cubes before 27 are 1, 8.

<prompt_3.1> Re-evaluate the problem to ensure 27 is the smallest possible cube that meets the criteria. </prompt_3.1>
-----

But wait, just to make sure, maybe there's a cube between 1 and 27 that can be expressed as the sum of three consecutive integers. But cubes between 1 and 27 are 8 and 27. Wait, 8 is the next cube after 1. But as we saw, 8 gives a non-integer n. Let me check again.

<prompt_3.2> Verify if there is a cube between 1 and 27 that can be expressed as the sum of three consecutive integers. </prompt_3.2>

But just to be thorough, let's check if there's another way to express a smaller cube as the sum of three consecutive integers, maybe if we consider negative integers? Wait, the problem says "smallest positive perfect cube", so we need the cube itself to be positive, but the integers can be negative.

<prompt_3.3> Consider if negative integers could result in a smaller positive perfect cube. </prompt_3.3>
-----

But just to make sure there isn't a smaller cube, let's check k=3 gives 27, which works. k=1 and k=2 don't work as they result in non-integer n. So 27 is the first one.

<prompt_3.4> Confirm that 27 is the smallest positive perfect cube that can be written as the sum of three consecutive integers. </prompt_3.4>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's see if 8 can be written as the sum of three consecutive integers. Suppose 8 = a + (a + 1) + (a + 2) = 3a + 3. Then 3a + 3 = 8 => 3a = 5 => a = 5/3. But that's not an integer. So, 8 is out. Similarly, 1 is even smaller, but 1 as a sum would require 3a + 3 = 1 => a negative fraction. So, no. Therefore, 27 is the first cube that works.
</execution_3.1>

<execution_3.2>
If \( k = 2 \), then the sum would be 8. Then the three consecutive integers would sum to 8. Let's see if that's possible. Let the three integers be \( n-1 \), \( n \), \( n+1 \). Their sum is \( 3n = 8 \), so \( n = 8/3 \), which is not an integer. So, that's not possible. 

So, indeed, the next cube is 27, which works.
</execution_3.2>

<execution_3.3>
For example, even if the three consecutive integers include negative numbers, their sum could still be a positive cube. But let's see.

Suppose someone tried to use negative integers. Let's see if that's possible. Let me take the same approach. Let the three consecutive integers be \( n-1 \), \( n \), \( n+1 \). Then their sum is \( 3n \). So, the sum is three times the middle number. Therefore, for the sum to be a positive perfect cube, \( 3n = k^3 \), so \( n = \frac{k^3}{3} \). So, \( n \) must be an integer, so \( k^3 \) must be divisible by 3, so \( k \) must be divisible by 3. Then, as before, the smallest \( k \) is 3, leading to \( n = 9 \), sum 27. So even if we consider centered around different middle numbers, the conclusion is the same. 

Alternatively, maybe there's another way? Suppose the three consecutive integers are not centered around the middle term. Wait, no, three consecutive integers must be centered around the middle term. So, regardless of how you denote them, the sum is three times the middle term. Therefore, the sum must be a multiple of 3. Therefore, the perfect cube must be a multiple of 3. Therefore, the cube must have 3 as a factor, so 3 divides \( k \), so \( k \) is a multiple of 3. Therefore, the smallest such cube is \( 3^3 = 27 \).
</execution_3.3>

<execution_3.4>
But let me check another angle. Suppose someone tried to use a different approach, maybe not starting with the middle term. Let's say the three consecutive integers are \( a-1 \), \( a \), \( a+1 \). Then their sum is \( 3a \). So, the sum must be a multiple of 3. Therefore, the perfect cube must be a multiple of 3. So, the cube must be divisible by 3, hence the cube must be divisible by 3^3=27 if the cube is to be minimal? Wait, no. Wait, if the cube is divisible by 3, then 3 divides the cube, so 3 divides the base, so the cube is divisible by 27. Therefore, the smallest such cube is 27. Because the next cube is 64, which is not divisible by 3. Then 125, also not divisible by 3. 216 is divisible by 27 (216=6^3=27*8). But 27 is smaller. So, 27 is the smallest cube divisible by 3, hence the sum of three consecutive integers. 

Therefore, the answer is 27.
</execution_3.4>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let's verify this. Let me check if there's another cube between 1 and 27 that can be expressed as such a sum. Wait, cubes between 1 and 27 are 8 (2^3) and 27 (3^3). We saw that 8 doesn't work, as the integers would be fractions.
So 27 is the first one.

But let me think again. Maybe there's a different way to represent the three consecutive integers? For example, if the three consecutive integers are not centered around the middle term. Wait, but three consecutive integers must be n-1, n, n+1, so the middle term is n. So, their sum is 3n.

<prompt_4.1> Re-evaluate the representation of consecutive integers and confirm the sum. </prompt_4.1>
-----

Based on execution_3.4:
But to be thorough, let's check for k=3, the cube is 27, sum of 8,9,10. Let's check if there's another set of three consecutive numbers that sum to 27. For example, 7,8,9: sum is 24. 8,9,10: 27. 9,10,11: 30. So, yes, 8,9,10 is the only triplet that sums to 27. 

Alternatively, maybe negative numbers? Let's see.

<prompt_4.2> Explore the possibility of negative integers and confirm that 27 is the smallest cube. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
So, regardless of how you denote them, the sum is 3n. Therefore, the sum must be a multiple of 3. So, the cube must be a multiple of 3, hence the cube must be 27, 216, etc. 

Therefore, 27 is the smallest such cube.
</execution_4.1>

<execution_4.2>
Suppose we have three consecutive integers including negative numbers. Let's see if their sum can be a positive cube smaller than 27. Let's try sum = 8. Let the three numbers be n-1, n, n+1. Then 3n = 8 => n = 8/3 ≈ 2.666. Not an integer. Similarly, for sum = 1: 3n =1 => n=1/3. Not integer. So, even if we allow negative integers, the same problem occurs. So, the middle term must be an integer, hence the sum must be a multiple of 3. Therefore, the cube must be a multiple of 3, hence the smallest such cube is 27. 

Therefore, I can confirm that 27 is the smallest positive perfect cube that can be written as the sum of three consecutive integers.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
**Final Answer**
The smallest positive perfect cube that can be written as the sum of three consecutive integers is \boxed{27}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the smallest positive perfect cube that can be written as the sum of three consecutive integers is $\boxed{27}$.
</Final_answer>
</think>