Proof of $AB=BA=O_n$ when $\ A^2+AB+B^2 = 2BA$

Question

Let n be a natural number and $\ A$,$\ B$ two real matrices of dimension 2 and at least one of them is invertible. Then if $\ A^2+AB+B^2 = 2BA$ prove that $AB=BA=O_n$. If I add $BA$ to the LHS then I have $\ (A+B)^2=3AB$ but if at least one is invertible,that is still not enough to prove we can do $\ a^3 + b^3=(\ a + b)(a^2 + ab + b^2)$. Any help?

Answer 1

Here is a counterexample. Take the matrices $A=I_2$ and $$ B=\begin{pmatrix} 1 & -1 \cr 1 & 0 \end{pmatrix}. $$ Then $A^2+AB+B^2=2BA$, because $B^2-B+I_2=0$, but $AB\neq 0$, and $BA\neq 0$.