Is the following true?
Let $A\in M_2(\mathbb R)$ and $A^3=I, A\neq I$,then $A^2+A+I=O$.
Solution: Yes the following statement is true.Let $m(x)$ be the minimal polynomial of $A$.Now $(x^3-1)$ annihilates $A$.So,$m(x)| x^3-1$ i.e. $(x^3-1)=q(x)m(x)$ .Now suppose $c\in \mathbb C$ is an eigenvalue of $A$,then $m(c)=0$ and hence $c^3=1$ which implies that $c=1$ or $\omega$ or $\omega^2$.Now,if $1$ is an eigenvalue of $A$,then $1$ is the only eigenvalue of $A$,because characteristic polynomial of $A$ has real coefficients and so imaginary roots must occur in conjugate pairs.So,$p(x)=(x-1)^2$ would be the characteristic polynomial of $A$ and hence by Cayley-Hamilton theorem $p(A)=O$ i.e. $A^2-2A+I=O$ and hence $O=A^2+A+I=3A\implies A=O$ but $A\neq O$ as $1$ is an eigenvalue.
So,we have a contradiction and hence $1$ cannot be an eigenvalue of $A$.
So,we are left with $\omega$ and $\omega^2$ and we know that since $p(x)$,the characteristic polynomial has real coefficients so, if one of them is an eigenvalue then other one is also.
So,$A$ has two distinct eigenvalues $\omega$ and $\omega^2$ and hence $p(x)=(x-\omega)(x-\omega^2)=x^2+x+1$ and so $A^2+A+I=O$.
Is my solution alright?
You are given that $$ 0 = A^3-I = (A^2+A+I)(A-I) $$ The assumption that $A-I\ne 0$ gives $x\ne 0$ such that $$(A^2+A+I)x=0.$$ Applying $A$ gives $$ (A^2+A+I)Ax = 0. $$ If $\{ Ax,x\}$ were a linearly-independent set, this set would be a basis of $\mathbb{R}^2$, which would imply that $A^2+A+I=0$. On the other hand, if $\{ Ax,x\}$ were not a linearly-independent set, there would exist $\lambda\in\mathbb{R}$ such that $Ax=\lambda x$; then $(A^2+A+I)x=0$ would force $\lambda^2+\lambda+1=0$, thereby giving the contradiction that $\lambda^2+\lambda+1=0$ has real solutions. So, $\{ Ax,x\}$ must be a basis of $\mathbb{R}^2$. Hence, $A^2+A+I=0$ because it is $0$ on this basis, which is the desired conclusion.