$P \neq Q.$

$(ii)$ $P^{3}=Q^{3}$ and $P^{2} Q=Q^{2} P$

To find The value of determinant of $P^{2}+Q^{2}.$

On subtracting the given equations, we get

${P^{3}-P^{2} Q=Q^{3}-Q^{2} P}$

$\Rightarrow$ ${P^{2}(P-Q)=Q^{2}(Q-P)}$

$\Rightarrow$ $(P-Q)\left(P^{2}+Q^{2}\right)=0$

Now, since $P \neq Q \quad[\text { given }]$

$\Rightarrow \quad P-Q \neq 0 \Rightarrow\left|P^{2}+Q^{2}\right|=0$

$\therefore $ $P^{2}+Q^{2}=0$