Proving divisibility of $a^3 - a$ by $6$ As part of a larger proof, I need to show why $a^3-a$ is always divisible by $6$. I'm having trouble getting started.

Consider the following options:


*

*$a\equiv0\pmod6 \implies a^3-a\equiv  0-0\equiv6\cdot 0\equiv0\pmod6$

*$a\equiv1\pmod6 \implies a^3-a\equiv  1-1\equiv6\cdot 0\equiv0\pmod6$

*$a\equiv2\pmod6 \implies a^3-a\equiv  8-2\equiv6\cdot 1\equiv0\pmod6$

*$a\equiv3\pmod6 \implies a^3-a\equiv 27-3\equiv6\cdot 4\equiv0\pmod6$

*$a\equiv4\pmod6 \implies a^3-a\equiv 64-4\equiv6\cdot10\equiv0\pmod6$

*$a\equiv5\pmod6 \implies a^3-a\equiv125-5\equiv6\cdot20\equiv0\pmod6$

