For all natural values of n, $n\left(n^2-1\right)$  is

$n-1,n,n+1$  are three consecutive natural numbers we have product of 3 consecutive natural numbers are always divisible by 2 and 3.

$\Rightarrow n\left(n^2-1\right)$  is a multiple of $\text{'}6\text{'}$