By factoring the expression, check that $n^3 - n$ is always divisible by 6 for all natural numbers $n$. Give reasons.

$\text{Factor: } n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1)$

This is the product of three consecutive integers.

Among any three consecutive integers, at least one is divisible by 2 (an even number), and exactly one is divisible by 3.

Therefore, the product is divisible by $2 \times 3 = 6$.
 Hence $n^3 - n$ is divisible by 6 for every natural number $n$.