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.

Assuming $A$ is invertible, the first relation implies
$B = A^{-1}(I+B^2A^2)$, then:
$BA -A^2 B^2=  A^{-1}(I+B^2A^2) A -A^2 B^2= I + A^{-1} B^2A^3  -A^2B^2= I - A^{-1} B^2B^3  -A^2B^2  =  I - A^{-1} B^3B^2  -A^2B^2 = I + A^{-1} A^3B^2  -A^2B^2 =  I + A^2B^2  -A^2B^2 = I$.
