Let P and Q be 3x3 matrices with P≠Q. If P³=Q³ and P²Q=Q²P, then determinant of (P²+Q²) is equal to:

Given that P and Q are 3x3 matrices such that P ≠ Q, P³ = Q³, and P²Q = Q²P.
We want to find the determinant of (P² + Q²).
Since P²Q = Q²P, we have P³ - Q³ = P(P²Q) - (Q²P)Q = P(Q²P) - Q(Q²P) = PQ²P - Q³P = P(P² - Q²)P = (P² - Q²)P² = 0
We have P³ - Q³ = 0, which means P³ = Q³.
Also, we have P³ - Q³ = (P - Q)(P² + PQ + Q²) = 0.
Since P ≠ Q, (P - Q) is not the zero matrix. Therefore, det(P - Q) ≠ 0 if and only if (P - Q) is invertible.
However, we have that (P - Q)(P² + PQ + Q²) = 0. If P - Q is invertible then (P² + PQ + Q²) must be the zero matrix.
But we are given that P²Q = Q²P. Then:
(P - Q)(P² + PQ + Q²) = P³ + P²Q + PQ² - QP² - Q²P - Q³ = P³ - Q³ + P²Q - Q²P + PQ² - QP² = 0
Since P³ = Q³, and P²Q = Q²P, we have PQ² - QP² = 0.
Let's consider the expression det(P² + Q²).
Let's assume det(P² + Q²) ≠ 0. Then P² + Q² is invertible.
We know that (P - Q)(P² + PQ + Q²) = 0. Since P³ = Q³, we have (P-Q)(P²+PQ+Q²) = 0.
If det(P - Q) ≠ 0, then P² + PQ + Q² = 0. However, this is not necessarily true. If det(P² + PQ + Q²) = 0, that doesn't mean det(P² + Q²) = 0.
Consider the case where P = Q. Then P³ = Q³ and P²Q = Q²P are trivially satisfied. In this case, det(P² + Q²) = det(2P²) = 2³det(P²) = 8(det(P))². This is not necessarily 0.
However, if P³ = Q³ and P²Q = Q²P, then P³ - Q³ = (P - Q)(P² + PQ + Q²) = 0. If P ≠ Q, then det(P² + PQ + Q²) = 0.
Multiplying by (P+Q), we get (P+Q)(P²+PQ+Q²)=P³+Q³+P²Q+PQ²+QP²+Q²P =2P³ + 2P²Q=0
If det(P-Q)=0 and P≠Q, then det(P²+Q²)=0
However, this is not necessarily true in general. The given condition is not enough to determine det(P²+Q²) definitively.