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

(N/A) We can factorize the expression $n^{3}-n$ as follows:
$n^{3}-n = n(n^{2}-1) = n(n-1)(n+1) = (n-1)n(n+1)$.
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 the product contains a factor of $2$ and a factor of $3$,the entire product must be divisible by $2 \times 3 = 6$.
Therefore,$n^{3}-n$ is divisible by $6$ for any positive integer $n$.