Show that $\operatorname{rank}(A^2+A+I_3)=1$ If $A \in M_3(\mathbb{R}), A \ne I_3 $ and $A^3=I_3$ Show that $\operatorname{rank}(A^2+A+I_3)=1$.   
What I have reached so far is that $\operatorname{rank}(A-I_3)+\operatorname{rank}(A^2+A+I_3)\le 3$ using Sylvester theorem, and also it can be quite easily proven that  $\operatorname{rank}(A^2+A+I_3) \ne 0$, if that helps in any way, but I have no idea what should I do now.
I've seen this problem statement in the archives of a contest, but no official solution is provided. 
 A: $A$ is a root of $x^{3}-1$, which has distinct roots, so $A$ can diagonalized over the complex numbers. Since it is a real matrix, and it is different from the identity, the diagonal form must be
$$B =\begin{bmatrix}
1&&\\
&\omega&\\
&&\omega^{2}
\end{bmatrix},$$
where $\omega$ is a primitive third root of unity. Since $1 + \omega + \omega^{2} = 0$, we have that
$$
B^{2} + B + I = \begin{bmatrix}
3&&\\
&0&\\
&&0
\end{bmatrix}
$$
has rank one. Now $A$ and $B$ are conjugate, and conjugacy preserves rank.
A: Since you know that $A^3=I$, we have $m_A(x)|x^3-1=(x-1)(x^2+x+1)$ (the minimal polynomial). Since the caracteristic polynomial and the minimal have the same irriducible factors, $m_A(x)\neq x^2+x+1$, since otherwise $p_A(x)=(x^2+x+1)^k$, but $\deg(p_A(x))=3$.
Since $A\neq I$, $m_A(x)\neq (x-1)$. Hence all your information is equivalent to $m_A(x)=x^3-1=(x-1)(x^2+x+1)$. Hence $A$ is diagonalizable over $\mathbb{C}$ to the form $D=diag(1,\alpha,\beta)$ where $\alpha,\beta$ are the roots of $x^2+x+1=0$.
Write $A=PDP^{-1}$. Then $A^2+A+I=P(D^2+D+I)P^{-1})$ and $rank(D^2+D+I)=1$ (you should see easily why). Hence $rank(A^2+A+I)=1$.
A: Hint: What can you say about the minimal polynomial of $A$?
