Given $A$ is a $3$rd order Real Symmetric Matrix such that $A^6=I$ Then Prove that $A^2=I$
My try: Given that $A^6=I$ we have
$$A^6-I=(A^2-I)(A^4+A^2+I)=0$$
Now our aim to prove $A^4+A^2+I$ is an invertible matrix
Also $$A^4+A^2+I=(A^2-A+I)(A^2+A+I)$$
So to prove both $A^2-A+I$ and $A^2+A+I$ both are invertible
Can i have any clue?