Prove if $A - I$ is invertible then $I + A + A^2 = (A^3 − I)(A − I)^{-1}$ Attempt at a solution:
If we look at the left side and multiply it by $(A - I)$, we get $AI + A^2 + A^3 - I^2 - IA - IA^2 = (A^3 - I)$. This is equal to $I(A^3 - I) = (A - I)(A - 1)^{-1}(A^3 - I)$. I just don't know what I would do to remove the $(A - I)$ and if you can switch the order to be $(A^3 - I)(A - I)^{-1} = I + A + A^2$ from there.

As you said, $(I+A+A^2)(A-I)=A^3-1$, 
so, if $A-I$ is invertible, 
then $I+A+A^2=(I+A+A^2)(A-I)(A-I)^{-1}=(A^3-1)(A-I)^{-1}$.
