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 !

Let $\lambda_i$ be eigenvalues of $A$. Given condition means that
$$
\lambda_1+\lambda_2+\lambda_3=0\quad\text{and}\quad\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}=0\quad\text{and}\quad\lambda_1\lambda_2\lambda_3=1
$$
Solve these and we have $\lambda_k=e^{k2\pi/3i}$. Since $A$ has distinct eigenvalues, it is similar to diagonal matrix, i.e
$$
P^{-1}AP=\pmatrix{\lambda_1 \\ & \lambda_2 \\ && \lambda_3}
$$
Thus
$$
A^3=P\pmatrix{\lambda_1^3 \\ & \lambda_2^3 \\ && \lambda_3^3}P^{-1}=PI_3P^{-1}=I_3
$$
