If a number n is whole number, which is 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 what `n^2(n^2 - 1)` is always divisible by when `n` is a whole number greater than 1.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We have 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 express our original expression as:
\[ n^2(n^2 - 1) = n^2(n - 1)(n + 1) \]

2. **Identifying the Factors**:
The expression now consists of three factors: `n^2`, `n - 1`, and `n + 1`. We need to analyze these factors to find common divisors.

3. **Analyzing the Factors**:
- `n^2` is always divisible by `n`.
- `n - 1` and `n + 1` are two consecutive integers. Therefore, one of them is always even.
- Since one of `n - 1` or `n + 1` is even, the product `n(n - 1)(n + 1)` will always include at least one factor of 2.

4. **Finding Divisibility by 3**:
Among any three consecutive integers (which includes `n - 1`, `n`, and `n + 1`), at least one of them must be divisible by 3. Therefore, the product will also be divisible by 3.

5. **Combining the Results**:
Since we have established that:
- The product is divisible by `n` (which is greater than 1).
- The product is divisible by 2 (from the even factor).
- The product is divisible by 3 (from the three consecutive integers).

We can conclude that the product `n^2(n^2 - 1)` is divisible by:
\[ 2 \times 3 = 6 \]

6. **Final Verification**:
To find a common divisor larger than 6, we can check specific values of `n`:
- For `n = 2`:
\[ n^2(n^2 - 1) = 4(4 - 1) = 4 \times 3 = 12 \]
- For `n = 3`:
\[ n^2(n^2 - 1) = 9(9 - 1) = 9 \times 8 = 72 \]
- For `n = 4`:
\[ n^2(n^2 - 1) = 16(16 - 1) = 16 \times 15 = 240 \]

All these results are divisible by 12.

### Conclusion:
Thus, we can conclude that `n^2(n^2 - 1)` is always divisible by **12** when `n` is a whole number greater than 1.