If n is any natural number then `n^(2)(n^(2)-1)` is always divisible

Options

A. 12
B. 24
C. `12-n`
D. 6

Answer

To solve the problem, we need to show that \( n^2(n^2 - 1) \) is always divisible by 12 for any natural number \( n \).

### Step-by-Step Solution:

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

2. **Identify the Factors**:
The expression \( n^2(n - 1)(n + 1) \) consists of three consecutive integers: \( n - 1 \), \( n \), and \( n + 1 \).

3. **Divisibility by 2**:
Among any three consecutive integers, at least one of them is even. Therefore, the product \( (n - 1)(n)(n + 1) \) is divisible by 2.

4. **Divisibility by 4**:
In addition to being even, among any three consecutive integers, at least one of them is divisible by 2. Therefore, we can guarantee that the product is divisible by \( 2 \times 2 = 4 \).

5. **Divisibility by 3**:
Among any three consecutive integers, at least one of them is divisible by 3. Thus, the product \( (n - 1)(n)(n + 1) \) is also divisible by 3.

6. **Combine the Factors**:
Now, we have established that:
- The product is divisible by 4 (from the even numbers).
- The product is divisible by 3 (from the three consecutive integers).

Therefore, the product \( n^2(n - 1)(n + 1) \) is divisible by:
\[ 4 \times 3 = 12 \]

### Conclusion:
Thus, we conclude that \( n^2(n^2 - 1) \) is always divisible by 12 for any natural number \( n \).