How can I prove that a square matrix is invertible if it satisfies this polynomial equation? For a 3x3 matrix $C$, it is given that
$$C^3+I=3C^2-C$$
I am then required to prove that $C$ is invertible.
I have attempted a proof, below, but I am not sure it is valid or if there is a better solution.
Attempted proof
$$C^3 + I = 3C^2 - C$$
$$I = - C^3 +3C^2-C$$
If it is assumed that $C^{-1}$ exists then
$$I = C^{-1}(-C^4+3C^3-C^2)$$
If $C^{-1}$ is defined, then $I=C^{-1}C$; therefore test whether
$$C \stackrel{!}{=} -C^4 + 3C^3 - C^2$$
$$ 0 = -C^4 + 3C^3 - C^2 - C$$
$$ C = 0, 1, 1\pm\sqrt{2}$$
Is this at all in the right direction?

Subtract to get $$I=-C^3+3C^2-C$$
Then factor to get $$I=C(-C^2+3C-I)$$
Now you  have $I=CD$, for $D=-C^2+3C-I$.  Hence $C$ is invertible.
