Let A and B be 3 × 3 matrices such that $A≠B$. If $A^3 = B^3$ and $A^2B = B^2A$, then the determinant of $A^2+B^2$ is:

0

The correct answer is Option (3) → 0

Given: $A^{3}=B^{3}$ and $A^{2}B=B^{2}A$

From $A^{3}=B^{3}$ ⇒ $(A-B)(A^{2}+AB+B^{2})=0$

Since $A \ne B$, the second factor must be singular:

$\Rightarrow \det(A^{2}+AB+B^{2})=0$

Also, $A^{2}B=B^{2}A \Rightarrow AB=BA$ (A and B commute)

Then $(A-B)(A^{2}+B^{2}+AB)=0$ gives $\det(A^{2}+B^{2}+AB)=0$

Because $A$ and $B$ commute, $AB$ behaves like a number term, so this implies:

$\det(A^{2}+B^{2})=0$