Prove $BA - A^2B^2 = I_n$. I have a problem with this. Actually, still don't have the right way to start :/
Problem :
Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$.
Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.
Thanks for any help.

The two given conditions $AB-B^2A^2=I$ and $A^3+B^3=0$ can be rewritten as
$$
\pmatrix{A&B^2\\ -B^2&A} \pmatrix{B&A^2\\ -A^2&B} = \pmatrix{I&0\\ 0&I}.
$$
Since $XY=I$ implies that $YX=I$ for any two square matrices $X$ and $Y$, we have
$$
\pmatrix{B&A^2\\ -A^2&B} \pmatrix{A&B^2\\ -B^2&A} = \pmatrix{I&0\\ 0&I}
$$
and the assertion follows.
