For any positive integer n,

Let $a=n^{3}-n \Rightarrow a=n \cdot\left(n^{2}-1\right)$

$\Rightarrow \quad a=n \cdot(n-1) \cdot(n+1)$ $\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]$

$\Rightarrow \quad a=(n-1) \cdot n \cdot(n+1)$ $\ldots$ (i)

We know that,

1. If a number is completely divisible by 2 and 3, then it is also divisible by 6.
2. If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
3. If one of the factor of any number is an even number, then it is also divisible by 2.

∴                                       a = (n – 1) . n . (n + 1)                                                 [from Eq. (i)]

Now, sum of the digits =n-1+n+n+1=3n

= multiple of 3, where n is any positive integer,

and (n -1)- n- (n + 1) will always be even, as one out of (n -1) or n or (n + 1) must of even. Since, conditions II and III is completely satisfy the Eq. (i).

Hence, by condition I the number n3 – n is always divisible by 6, where n is any positive

integer.                                                                                                               Hence proved.