In $\mathbb{Z}_6$, $3^3 = 3^{-3}$ since $3^{-3} = 3^{6-3} = 3^3$. Thus $(3)^3 = (3^{-1})^3=2^3=2$. But also $3^3 = 3$ in $\mathbb{Z}_6$. Where's my error? Sorry for this question, but I think I got too confused and better ask for help.
Resolved: I mixed up the additive and multiplicative groups in my head. Need more rest.
What could you possibly mean with $3^{-1}$ (by extension $3^{-3}$)? In $\Bbb{Z}/6\Bbb{Z}$, $3$ isn't invertible.