The greatest possible number which can always divide the sum of the cubes of any three consecutive integer is:

Options

A. 3
B. 4
C. 9
D. 11

Answer

To find the greatest possible number that can always divide the sum of the cubes of any three consecutive integers, we can follow these steps:

### Step-by-Step Solution:

1. **Define the three consecutive integers**: Let the three consecutive integers be \( n \), \( n+1 \), and \( n+2 \).

2. **Calculate the cubes of these integers**:
- The cube of \( n \) is \( n^3 \).
- The cube of \( n+1 \) is \( (n+1)^3 = n^3 + 3n^2 + 3n + 1 \).
- The cube of \( n+2 \) is \( (n+2)^3 = n^3 + 6n^2 + 12n + 8 \).

3. **Sum the cubes**:
\[ \text{Sum} = n^3 + (n^3 + 3n^2 + 3n + 1) + (n^3 + 6n^2 + 12n + 8) \]
Simplifying this, we get:
\[ \text{Sum} = 3n^3 + (3n^2 + 6n^2) + (3n + 12n) + (1 + 8) \]
\[ \text{Sum} = 3n^3 + 9n^2 + 15n + 9 \]

4. **Factor the sum**:
We can factor out a 3 from the entire expression:
\[ \text{Sum} = 3(n^3 + 3n^2 + 5n + 3) \]

5. **Determine the divisibility**:
The expression \( n^3 + 3n^2 + 5n + 3 \) can take various values depending on \( n \), but the factor of 3 is constant. Therefore, the sum of the cubes of any three consecutive integers is always divisible by 3.

6. **Check for greater divisibility**:
To find if there is a greater number that can always divide this sum, we can evaluate the expression \( n^3 + 3n^2 + 5n + 3 \) for different values of \( n \) to see if it is divisible by any number greater than 3.

- For \( n = 1 \):
\[ 1^3 + 3(1^2) + 5(1) + 3 = 1 + 3 + 5 + 3 = 12 \quad (\text{divisible by } 4) \]
- For \( n = 2 \):
\[ 2^3 + 3(2^2) + 5(2) + 3 = 8 + 12 + 10 + 3 = 33 \quad (\text{not divisible by } 4) \]

Since the expression is not consistently divisible by 4, we conclude that 3 is the greatest number that can always divide the sum of the cubes of any three consecutive integers.

### Final Answer:
The greatest possible number which can always divide the sum of the cubes of any three consecutive integers is **3**.