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? 

Is it not enough to check if this polynomial has a double root?
Its derivative is
$$
8x^7 + 2x = x(8x^6 + 2) = 8x(x^6 + \frac 14).
$$
Now $0$ is not a common root, so the double roots would satisfy
$$
x^6 + \frac14 = 0.
$$
Plugging this into our original polynomial, a double root would satisfy
$$
0 = x^8 + x^2 - 1 = x^2(x^6 + \frac14) + \frac34 x^2 -1= \frac34 x^2 -1,
$$
but the roots of $x^2 - \frac43$ are not roots of $x^6 + \frac14$,
since if $x^2 = \frac 43$, then
$$
x^6 + \frac14 = \left(\frac43\right)^3 + \frac14 \neq 0.
$$
