If $a\in\Bbb{Z}$ then $a^3\equiv a\pmod{3}$.
For the sake of contradiction, assume $a\in\Bbb{Z}$ and $a^3\not\equiv a\pmod{3}$. Then $3\not\mid (a^3-a)$ which implies that either $(a^3-a)=3q+1$ or $(a^3-a)=3q+2$.
Take the case where $(a^3-a)=3q+1$. Then either $q$ is even or odd. If we assume $q$ to be even, then we have $(a^3-a)=3(2n)+1, n\in\Bbb{Z}$ which implies that $(a^3-a)=2(3n)+1$ and therefore $(a^3-a)$ is odd.
But, factoring gives us $(a^3-a)=a(a+1)(a-1)$. Then either $a$ is even or $a$ is odd. But both of these cases show that $(a^3-a)$ would be even. Therefore we have a contradiction.
$\blacksquare$
I'm new to writing proofs so I'm unclear if this single contradiction is sufficient or if it is too narrow of an example.
For your proof to work, you’d still need to discard the case where $a^3-a$ is of the form $3k+2$ – the argument will be almost identical. However, your observation that $$a^3-a=a(a+1)(a-1)$$ immediately implies that this expression is always a multiple of three, without needing to resort to a contradiction (since one of three consecutive integers is always a multiple of three). This was exactly what you wanted to prove.