What will be the value of $\det(A^2+B^2)$?

Question

Let $A$ and $B$ be two $n\times n$ matrices such that $A \not=B, A^3 = B^3$ and $A^2 B = B^2 A$. Then what is the value of $\det \left( A^2 + B^2 \right)$?

Here I have done something.

$$A^3 = B^3 \implies(A-B)(A^2+B^2+AB)=0$$

Is my work right? Then how to do further? Well, I did mistakes here. But how can find the value of $\det(A^2+B^2)$?

Answer 1

the following observation may help:

set $X=A^2+B^2$

then $$ XA=A^3+B^2A = B^3 + A^2B = XB $$ so $$ X(A-B) = 0 $$

Answer 2

@ my stak , to your question "is my work right ?", the answer is: no, it is catastrophic; in addition, you removed the most hilarious part of your work.

Since $A-B\not=0$, there is $u$ s.t. $v=(A-B)u\not=0$. According to the David's work, $Xv=0$ and $\det(X)=0$.