Here is an exercise that I have been used in a oral exam (around 45 minutes) for undergrad students.
Let $A$ a $n\times n$ matrix with real coefficients such that $$ A^T = 3A^2-A-I_n,$$ where $A^T$ is the transpose matrix of A and $I_n$ the identity matrix of size $n\times n$.
Here is what I have found: We have $$ A = 3(A^T)^2-A^T-I_n,$$ and using that we have $$ A = 3(3A^2-A-I_n)^2-(3A^2-A-I_n)-I_n.$$ Expanding the previous identity, we have $$ 3I_n+6A-18 A^2-18 A^3+27 A^4=3 (3 A+1) (A-1) (3 A^2-1)=0.$$ So that $1$ is a possible eigenvalue of $A$, which can be difficult to deal with when we look at powers of a matrix. Consequently, I decide to write $$ X^n = (3 X+1) (X-1) (3 X^2-1)Q(X)+R(X),$$ withe $\deg R < 4$. One can find the exact formula for $R$ (which is ugly (in the basis $(1,X,X^2,X^3)$) see here) and see that $A^n = R(A)$ converge.
Is there an easiest way to solve this exercise?
You have already shown that $ (3A+1)(A-1)(3A^2-1) = 0$, so the only possible eigenvalues for $A$ are $-\frac{1}{3}$, $1$, and $\pm \frac{1}{\sqrt{3}}$. They all have their module $<1$, except $1$.
Since this polynomial cancels $A$ and it has simple roots, it implies that you can diagonalize $A$. If you do so, you will get a diagonal matrix with the eigenvalues on the diagonal. This diagonal matrix easily converges, because $\lambda^n$ will converge to 0 if $\vert \lambda \vert <1$, and to 1 if $\lambda = 1$, when $n \rightarrow \infty$.