Let A and B be 3×3 matrices such that $A \neq B$. If $A^3 = B^3$ and $A^2B = B^2A$, then the determinant of $A^2 + B^2$ is:

Given: $A$ and $B$ are $3 \times 3$ matrices such that $A \neq B$, $A^3 = B^3$, and $A^2B = B^2A$.

We need to find $|A^2 + B^2|$.

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We know:

$A^3 = B^3$ ... (1)

$A^2B = B^2A$ ... (2)

Subtracting (2) from (1):

$A^3 - A^2B = B^3 - B^2A$

Bringing everything to one side:

$A^3 - A^2B + B^2A - B^3 = 0$

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Taking $A^2$ common from the first two terms and $B^2$ common from the last two terms:

$A^2(A - B) + B^2(A - B) = 0$

Now $(A - B)$ is common on the right side of both terms, so:

$(A^2 + B^2)(A - B) = 0$ ... (3)

(In matrices, order matters. Here $(A - B)$ is on the right in both terms, so taking it out on the right is correct.)

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Let us assume $|A^2 + B^2| \neq 0$.

Then $(A^2 + B^2)^{-1}$ exists.

Multiplying both sides of (3) on the left by $(A^2 + B^2)^{-1}$:

$(A^2 + B^2)^{-1}(A^2 + B^2)(A - B) = 0$

$I(A - B) = 0$

$A = B$

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But we are given that $A \neq B$.

So our assumption is wrong, and $(A^2 + B^2)$ is not invertible.

A matrix that is not invertible has determinant zero.

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Therefore:

$|A^2 + B^2| = 0$