How to prove that for all integers $3|x^3-x$ I'm not sure how to prove that the following statement is true.
$\forall x \in \mathbb{Z}, 3|(x^3-x)$
$x(x-1)(x+1) = 3n $ where $n\in \mathbb{Z}$

By Fermat's Little Theorem, $$x^3\equiv x\pmod 3$$
since $3$ is prime. Hence,
$$x^3-x\equiv 0\pmod 3\iff 3\mid x^3-x$$

Or, more in line with what you've started, we have $x^3-x=x(x-1)(x+1)$. Either $x,x-1$ or $x+1$ is divisible by $3$ for any $x\in\Bbb Z$. Indeed, suppose $x$ is not divisible by $3$, then $x+1$ or $x+2$ is. In the former case, we have $3\mid x+1$, and in the latter case, we have $3\mid x+2\iff 3\mid x-1$.
