Proving the following matrix is diagonalizable I'm asked to prove that the matrix $A\in M_{n}(\mathbb C)$ that satisfy $A^8+A^2=I$ is diagonalizable. I've tried looking at the equation $x^8+x^2-1=0$ and determining whether $M_A$ has any repeating roots, but this got me nowhere. Afterwards, I thought about trying to determine whether its Jordan form is diagonal (I know such form exists since $\mathbb C$ is algebraically closed, so $P_A$ splits into linear factors) still got nowhere. 
Is there a right approach to the question or is there something I'm missing? 

Look at $g(x)=x^4+x-1$ . Observe that $g'(x)=4x^3+1$ has only one real root and deduce that g as no repeated root in $\mathbb R$.
Moreover, we have $g(0)=-1$ and $g(x)\to \infty$ for $x\to \infty$ and $x\to -\infty$.
So $g$ has only two distinct real roots. The other two roots are distinct and complex occurring in conjugate pairs.
Now we have $f(x)=x^8+x^2-1=g(x^2)$
$\therefore \{\alpha \in \mathbb C|f(x)=0\}=\{\alpha \in \mathbb C|\alpha^2 \text{ is a root of }g\} $
As the roots of $g$ are distict so are the roots of $f$.
So $f$ is product of distinct linear factors.
Now you know your minimal polynomial $M_A$ divides $f$ in $\mathbb C[x]$ as $f$ annihilates $A$. 
Hence $M_A$ is just the product of distinct linear factors hence $A$ is diagonalizible.  
