TITLE: Show that $\operatorname{rank}(A^2+A+I_3)=1$
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: Hint: What can you say about the minimal polynomial of $A$?