Key Concept: Properties of Matrix Multiplication and Determinants

This problem hinges on a crucial property of matrix algebra: If the product of two matrices, $A$ and $B$, is the zero matrix (i.e., $AB = 0$), and if $B$ itself is a non-zero matrix, then matrix $A$ must be a singular (non-invertible) matrix. A singular matrix, by definition, has a determinant of zero. This is a common point of distinction between scalar and matrix algebra, where for scalars $ab=0$ and $b \ne 0$ implies $a=0$. For matrices, $A$ doesn't necessarily have to be the zero matrix, but it must be non-invertible.

Given Information

We are given two $3 \times 3$ matrices, $P$ and $Q$, with the following conditions:

1. $P \ne Q$ (This implies that the matrix $(P-Q)$ is not the zero matrix).
2. $P^3 = Q^3$ --- (Equation 1)
3. $P^2Q = Q^2P$ --- (Equation 2)

Our goal is to find the determinant of the matrix $(P^2 + Q^2)$.

Step-by-Step Solution

Step 1: Manipulating the Given Equations

The first step is to combine the given equations in a way that allows for factorization. Notice that both equations involve powers of $P$ and $Q$. Subtracting Equation (2) from Equation (1) is a strategic move because it creates terms that can be grouped and factored.

Subtract Equation (2) from Equation (1): $P^3 - P^2Q = Q^3 - Q^2P$

Now, we want to bring all terms to one side to facilitate factorization. Moving all terms to the left-hand side gives: $P^3 - P^2Q - Q^3 + Q^2P = 0$ (Note: '0' here represents the $3 \times 3$ zero matrix, as $P$ and $Q$ are $3 \times 3$ matrices.)

Step 2: Factorization

Next, we group terms to identify common factors. We aim to factor out $(P-Q)$ from the expression. Let's group the terms: $(P^3 - P^2Q) + (Q^2P - Q^3) = 0$

Now, we can factor out $P^2$ from the first group and $Q^2$ from the second group. Remember that matrix multiplication is not commutative in general, so the order of factors matters. From the first group, $P^3 - P^2Q = P^2(P - Q)$. Here, $P^2$ is a common left factor. From the second group, $Q^2P - Q^3 = Q^2(P - Q)$. Here, $Q^2$ is also a common left factor.

Substituting these back into the equation: $P^2(P - Q) + Q^2(P - Q) = 0$

Now, observe that $(P-Q)$ is a common right factor in both terms. Using the distributive property for matrices, we can factor out $(P-Q)$: $(P^2 + Q^2)(P - Q) = 0$

This is a critical intermediate result. Let's call $A = (P^2 + Q^2)$ and $B = (P - Q)$. So we have $AB = 0$.

Step 3: Utilizing the Condition $P \ne Q$

We have the matrix equation $(P^2 + Q^2)(P - Q) = 0$. We are given that $P \ne Q$. This condition is crucial, as it means the matrix $(P - Q)$ is not the zero matrix. In other words, $B \ne 0$.

Now we apply the key concept discussed at the beginning: If $A$ and $B$ are matrices such that $AB = 0$ and $B \ne 0$, then $A$ must be a singular (non-invertible) matrix.

Let's prove this briefly: Assume, for the sake of contradiction, that $A$