If n is a whole number greater than 1, then `n^2 (n^2- 1)` is always divisible by :

Options

A. 16
B. 12
C. 10
D. 8

Answer

To solve the problem, we need to determine if \( n^2(n^2 - 1) \) is always divisible by a certain number when \( n \) is a whole number greater than 1.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( n^2(n^2 - 1) \). We can rewrite \( n^2 - 1 \) using the difference of squares:
\[ n^2 - 1 = (n - 1)(n + 1) \]
Therefore, we can rewrite the original expression as:
\[ n^2(n^2 - 1) = n^2(n - 1)(n + 1) \]

2. **Identifying the Factors**:
The expression now consists of four factors: \( n^2 \), \( n - 1 \), \( n + 1 \). Notice that \( n - 1 \), \( n \), and \( n + 1 \) are three consecutive integers.

3. **Properties of Consecutive Integers**:
Among any three consecutive integers, at least one of them is divisible by 2 (even), and at least one of them is divisible by 3. This means:
- There are at least two factors of 2 (since at least one of the three numbers is even).
- There is at least one factor of 3.

4. **Calculating the Divisibility**:
From the factors identified:
- The contribution from the two even numbers gives us \( 2 \times 2 = 4 \) (from two factors of 2).
- The contribution from the factor of 3 gives us \( 3 \).

Therefore, the product of these contributions is:
\[ 4 \times 3 = 12 \]

5. **Conclusion**:
Since \( n^2(n^2 - 1) \) is always divisible by 12 for any whole number \( n > 1 \), we conclude that the answer to the question is:
\[ \text{The expression is always divisible by } 12. \]