If $A \in M_3(\mathbb C)$ and $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$, then $A^3=I_3$

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 $$ A^3=P\pmatrix{\lambda_1^3 \\ & \lambda_2^3 \\ && \lambda_3^3}P^{-1}=PI_3P^{-1}=I_3 $$

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.