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: Expand the expression $(1+1)^3$.  
