<s>[INST] $A^2 - B^2$ is an invertible matrix and $A^5 = B^5$ and $A^3B^2 = A^2B^3$ . Then what is the determinant of the matrix $A^3 + B^3$? $A,B$ are two $3 \times 3$ real-matrices with following three properties.

*

*$A^2 - B^2$ is an invertible matrix

*$A^5 = B^5$

*$A^3B^2 = A^2B^3$
Then what is the determinant of the matrix $A^3 + B^3$ ?
 [/INST] $A$ and $B$ are real $3\times3$ matrices, so their eigenvalue polynomials are cubic and hence have at least one real zero. Let $\lambda\in\Bbb{R}$ be an eigenvalue of $B$, and $x\in\Bbb{R}^3$ be the corresponding eigenvector. Then
$$
A^5x=B^5x=\lambda^5x.\tag{1}
$$
We also have
$$
\lambda^3A^2x=(A^2B^3)x=A^3B^2x=\lambda^2A^3x.\tag{2}
$$
Apply $A^3$ to both sides of $(2)$, and use $(1)$ to arrive at
$$
\lambda^8x=A^3(\lambda^3A^2x)=A^3(\lambda^2A^3x)=\lambda^2A^6x=\lambda^7Ax.\tag{3}
$$
If $\lambda\neq0$ then $(3)$ implies that $Ax=\lambda x$ also. This violates the first assumption as then
$$(A^2-B^2)x=\lambda^2x-\lambda^2x=0.$$
We are left with the case $\lambda=0$. Again, by $(1)$ we see that $A^5x=0$. Because $A$ is a $3\times3$ matrix this implies (think Jordan canonical form) that already $A^3x=0$.
We have shown that $(A^3+B^3)x=0$ for a non-zero vector $x$, and can conclude that $\det(A^3+B^3)=0$.
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