The sum of the cubes of three consecutive natural numbers is divisible by

Options

A. 7
B. 9
C. 25
D. 26

Answer

To solve the problem of finding the divisibility of the sum of the cubes of three consecutive natural numbers, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Consecutive Natural Numbers**:
Let the three consecutive natural numbers be \( n-1, n, n+1 \), where \( n \) is a natural number.

2. **Write the Expression for the Sum of Their Cubes**:
We need to find the sum of their cubes:
\[ (n-1)^3 + n^3 + (n+1)^3 \]

3. **Expand Each Cube**:
Using the binomial expansion:
- \( (n-1)^3 = n^3 - 3n^2 + 3n - 1 \)
- \( n^3 = n^3 \)
- \( (n+1)^3 = n^3 + 3n^2 + 3n + 1 \)

4. **Combine the Expanded Terms**:
Now, we can add these three expansions together:
\[ (n^3 - 3n^2 + 3n - 1) + n^3 + (n^3 + 3n^2 + 3n + 1) \]

5. **Simplify the Expression**:
Combining like terms, we have:
\[ n^3 + n^3 + n^3 - 3n^2 + 3n + 3n + 1 - 1 = 3n^3 + 6n \]

6. **Factor the Result**:
We can factor out \( 3n \):
\[ 3n(n^2 + 2) \]

7. **Analyze the Factors**:
The expression \( 3n(n^2 + 2) \) shows that the sum is divisible by \( 3 \). Now, we need to check if it is divisible by \( 9 \).

8. **Check Divisibility by 9**:
- If \( n \) is a multiple of \( 3 \), then \( 3n \) is divisible by \( 9 \).
- If \( n \) is not a multiple of \( 3 \), then \( n \) can be expressed as \( 3k + 1 \) or \( 3k + 2 \). In both cases, \( n^2 + 2 \) will yield a result that is divisible by \( 3 \) (since \( (3k + 1)^2 + 2 \equiv 1 + 2 \equiv 0 \mod 3 \) and \( (3k + 2)^2 + 2 \equiv 1 + 2 \equiv 0 \mod 3 \)). Therefore, \( 3n(n^2 + 2) \) will also be divisible by \( 9 \).

### Conclusion:
Thus, we conclude that the sum of the cubes of three consecutive natural numbers is always divisible by \( 9 \).