For any positive integer $n,$ prove that $n^{3}-n$ is divisible by $6.$

(N/A) Let $a = n^{3} - n.$
Factorizing the expression,we get $a = n(n^{2} - 1).$
Using the identity $a^{2} - b^{2} = (a - b)(a + b),$ we have $a = (n - 1)n(n + 1).$
This expression represents the product of three consecutive integers.
In any set of three consecutive integers,at least one must be even (divisible by $2$) and exactly one must be divisible by $3.$
Since $2$ and $3$ are coprime,their product $2 \times 3 = 6$ must divide the product of these three consecutive integers.
Therefore,$(n - 1)n(n + 1)$ is always divisible by $6$ for any positive integer $n.$
Hence,$n^{3} - n$ is divisible by $6.$