I also remember that the determinant has some nice properties, such as $\det (\mathbf{AB}) = \det (\mathbf{A}) \det (\mathbf{B})$ for any square matrices $\mathbf{A}$ and $\mathbf{B}.$ Therefore, I can apply this property to the matrix $\mathbf{A^3},$ which is the same as $\mathbf{A \cdot A \cdot A}.$
[* { box }] I get $\det (\mathbf{A^3}) = \det (\mathbf{A \cdot A \cdot A}) = \det (\mathbf{A}) \det (\mathbf{A}) \det (\mathbf{A}) = \det (\mathbf{A})^3.$


If $\det \mathbf{A} = 5,$ then find $\det (\mathbf{A^3}).$