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 $n\in \mathbb{Z}$
 A: 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 $3\mid x+2\iff 3\mid x-1$.
A: $$x=3k+0\\x=3k+1\\x=3k+2$$ $$3|(x-1)x(x+1)$$
$$x=3k  \to 3|(3k-1)(3k)(3k+1) \checkmark$$
$$x=3k+1  \to 3|(3k+1-1)(3k+1)(3k+1+1) \checkmark$$
$$x=3k+2 \to 3|(3k+2-1)(3k+2)(3k+2+1)\\3|(3k+2-1)(3k+2)(3(k+1)) \checkmark$$
A: Some great answers already, so I'm going to show you a more laborious way to get at the same result.
What are the cubes modulo 3? $$0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, \ldots$$ As you can see, this boils down to three cases.


*

*If $x$ is a multiple of 3 to begin with, then so is $x^3$ (that's a multiple of 27, in fact), and so is $x^3 - x$.

*If $x \equiv 1 \pmod 3$, then $x^3 \equiv 1 \pmod 3$ also. Then $x^3 - x \equiv 1 - 1 = 0 \pmod 3$.

*If $x \equiv 2 \pmod 3$, then $x^3 \equiv 2 \pmod 3$ also. Then $x^3 - x \equiv 2 - 2 = 0 \pmod 3$.

