Prove that $a^3 − a \equiv 0 \pmod 3$.
Using the Division Algorithm, we'll write $a$ as
$$ a = 3q + r $$
for integers $q,r$ where $0 \le r < 3$.
There are 3 cases for $r$, $r=0$, $r=1$, and $r=2$. Let's consider each in turn,
For $r=0$,
$$ \begin{align} a^3 - a & = 3^3q^3 - 3q \\ \\ & = 3(3^2q^3 - q) \\ \\ & \equiv 0 \pmod {3} \end{align} $$
For $r=1$,
$$ \begin{align} a^3 - a & = (3q+1)^3 - (3q+1) \\ \\ &= 27 q^3 + 27 q^2 + 6 q \\ \\ & = 3(9q^3 + 9q^2 + 2q) \\ \\ & \equiv 0 \pmod {3} \end{align} $$
For $r=2$,
$$ \begin{align} a^3 - a & = (3q+2)^3 - (3q+2) \\ \\ & = 27 q^3 + 54 q^2 + 33 q + 6 \\ \\ & = 3(9 q^3 + 18 q^2 + 11 q + 2) \\ \\ & \equiv 0 \pmod {3} \end{align} $$
So for all cases of $r$, we have $a^3 − a \equiv 0 \pmod 3$.