Let A and B be two 3 times 3 real matrices su | vedclass

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(A) Given that $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$.Subtracting the second equation from the first,we get:$A^{5}-A^{3} B^{2} = B^{5}-A^{2} B^{3

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(A) Given that $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$.Subtracting the second equation from the first,we get:$A^{5}-A^{3} B^{2} = B^{5}-A^{2} B^{3}$$A^{3}(A^{2}-B^{2}) = B^{5}-A^{2} B^{3}$Rearranging the terms:$A^{3}(A^{2}-B^{2}) + B^{3}(A^{2}-B^{2}) = 0$$(A^{3}+B^{3})(A^{2}-B^{2}) = 0$Since $(A^{2}-B^{2})$ is an invertible matrix,we can post-multiply by its inverse $(A^{2}-B^{2})^{-1}$:$(A^{3}+B^{3})(A^{2}-B^{2})(A^{2}-B^{2})^{-1} = 0 \cdot (A^{2}-B^{2})^{-1}$$A^{3}+B^{3} = 0$Therefore,the determinant of the zero matrix is $|A^{3}+B^{3}| = |0| = 0$.

Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2}-B^{2})$ is an invertible matrix. If $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$,then the value of the determinant of the matrix $A^{3}+B^{3}$ is equal to:

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