Eigenvalues of the cubed inverse of A minus 3I

The eigenvalues of an $n\times n$ matrix $A$ are $5$ and $7$. Let $B=\left((A-3I_n)^{-1}\right)^{3}.$ Add up the distinct eigenvalues of $B$.

$9/64$. If $Av=\lambda v$ then $Bv=(\lambda-3)^{-3}v$, so the eigenvalues of $B$ are $2^{-3}=\tfrac18$ and $4^{-3}=\tfrac1{64}$, whose sum is $\tfrac{9}{64}$.