Ring where $s^3-s=0$. Show $6=0$. Let $S$ be a ring in which every element $s$ satisfies $s^3-s=0$.
Show that $6=0$ in $S$.
I'm not sure if I should relate this to a concrete example involving integers, or attempt to prove it completely abstractly. I believe there is a proof by Jacobson that such a ring is commutative. I'm not sure if this helps us.

Hint:$s^3-s=s(s-1)(s+1)=0, s=2$, $2(2-1)(2+1)=6=0$.
