$A$ is diagonalizable if $A^8+A^2=I$ Given a matrix $A\in M_{n}(\mathbb{C})$ such that $A^8+A^2=I$, prove that $A$ is diagonalizable.
So let $p(x)=x^8+x^2-1$ and we know that $p(A)=0$.
The next step would be to show that the algebric and geometric multipliciteis of all the eigenvalues are equal.
But this polynomial is reducible in a very unpleasent way, so even checking for the minimal polynomial is not an option.
What can I do differently.

It suffices to show that all the eigenvalues are simple. If $\lambda$ is an eigenvalue with multiplicity $\ge2$ then we have
$$\lambda^8+\lambda^2-1=0\tag 1$$
and
$$8\lambda^7+2\lambda=0$$
but clearly $0$ isn't an eigenvalue so
$$\lambda^6=-\frac14\tag2$$
so form $(1)$ we get $$-\frac14\lambda^2+\lambda^2-1=\frac34\lambda^2-1=0\iff \lambda=\pm  \sqrt{\frac{4}{3}}$$
which contradicts $(2)$. Conclude.
