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

We are given two $$3 \times 3$$ matrices $$P,Q$$ such that $$P \neq Q$$, $$P^{3}=Q^{3}$$ and $$P^{2}Q = Q^{2}P$$. We have to find $$\det(P^{2}+Q^{2})$$.

Step 1 : Form the product $$(P^{2}+Q^{2})(P-Q)$$.
Expand it term-by-term:

$$\begin{aligned} (P^{2}+Q^{2})(P-Q) &= P^{2}P - P^{2}Q + Q^{2}P - Q^{2}Q \\ &= P^{3} - P^{2}Q + Q^{2}P - Q^{3}. \end{aligned}$$

Step 2 : Use the given relations.
We know $$P^{3}=Q^{3}$$ and $$P^{2}Q = Q^{2}P$$. Substituting these into the expansion gives

$$P^{3} - P^{2}Q + Q^{2}P - Q^{3} \;=\; 0 - 0 \;=\; 0,$$

so we have obtained the matrix equation

$$\bigl(P^{2}+Q^{2}\bigr)(P-Q) = 0 \quad -(1)$$

Step 3 : Draw the consequence for the determinant.
Suppose, for contradiction, that $$\det(P^{2}+Q^{2}) \neq 0$$. Then $$P^{2}+Q^{2}$$ is invertible, and we can premultiply both sides of $$(1)$$ by its inverse:

$$P-Q = (P^{2}+Q^{2})^{-1}\,0 = 0.$$

This would imply $$P=Q,$$ which contradicts the given condition $$P \neq Q$$. Therefore our supposition is impossible, and

$$\det(P^{2}+Q^{2}) = 0.$$

Hence the correct option is
Option C which is: $$0$$.