If $A \in M_3(\mathbb C)$ and $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$, then $A^3=I_3$ Assume that we have a $3\times 3$ matrix like $A$ in which the coordinates come from the set of complex numbers ($\mathbb C$).
We know that $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$ .
Prove that $A^3=I_3$ .  
I don't know what's the relationship between $A^3$ and the trace !

Expanding on the Hint of Ilisaryus: Use that the trace is the sum of the eigenvalues and the determinant the product and that the eigenvalues of $A^{-1}$ are the inverses of the eigenvalues of $A$. Find the characteristic polynomial and use Cayley-Hamilton.
