Let  $PandQ$  be  $3\times 3$  matrices $P\ne Q$ .  If $P^3=Q^3,\det (P-Q)\ne 0and{P}^{2}Q=Q^{2}P,$   then determinant of  $(P^2+Q^2)$  is equal to

$P^3=Q^3\text{ and }{P}^{2}Q=Q^{2}P$

$\begin{array}{r}{P}^3-P^{2}Q=Q^3-Q^{2}P\\ \Rightarrow P^2(P-Q)=-Q^2(P-Q)\\ \Rightarrow \left(P^2+Q^2\right)(P-Q)=0\end{array}$

$\text{ If det }\left(P^2+Q^2\right)\ne 0\text{ then }{P}^2+Q^2\text{ is invertible and hence }\mathrm{P}=Q\text{. Therefore, }\det \left(P^2+Q^2\right)=0$