If n be any natural number then by which largest number `(n^3-n)` is always divisible? 3 (b) 6 (c) 12 (d) 18

Explanation

To determine the largest number by which \( n^3 - n \) is always divisible for any natural number \( n \), we can follow these steps:

### Step 1: Factor the Expression
First, we can factor the expression \( n^3 - n \):
\[ n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1) \]
This shows that \( n^3 - n \) is the product of three consecutive integers: \( n \), \( n - 1 \), and \( n + 1 \).
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