Let $P$ and $Q$ be $3\times 3$ matrices with $P\ne Q$. If $P^3=Q^3$ and $P^{2}Q=Q^{2}P$, then determinant of $(P^2+Q^2)$ is equal to:

We have $P^3-Q^3=0$ and $P^{2}Q-Q^{2}P=0$.Subtracting the equations:$P^3-P^{2}Q-Q^3+Q^{2}P=0$$P^2(P-Q)+Q^2(P-Q)=0$$(P^2+Q^2)(P-Q)=0$ If $\det (P^2+Q^2)\ne 0$, then $(P^2+Q^2)$ would be invertible.Multiplying by $(P^2+Q^2{)}^{-1}$ from the left:$(P-Q)=0⟹P=Q$.But the problem states $P\ne Q$.Therefore, $\det (P^2+Q^2)$ must be equal to $0$.