Proving complex normal matrix is self-adjoint

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Let $A\in M_{n\times n}^{\mathbb{C}}$ be a normal complex matrix that satisfies $A^{9}+3A^{5}=3A^{7}+A^{3}$.

I need to prove that $A$ is self-adjoint (Hermitian) and also $A^{2}=A^{4}$.

It's easy to see that if $p(t)=(t-1)^{3}$ then $p(A^{2})=O$. Could it be the minimal/characteristic polynomial of $A^2$? If it is, how so? I figure that finding the ch.polynomial of $A$ and proving that all its roots are real is sufficiant for proving this.

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If $\lambda$ is an eigenvalue of $A$, then $$ \lambda^9-3\lambda^7+3\lambda^5-\lambda^3=0 $$ (why?) so $$ \lambda^3(\lambda^2-1)^3=0 $$ Therefore all eigenvalues of $A$ are real, which is an equivalent condition for a normal matrix to be self-adjoint.

The second part uses the fact that $\lambda\in\{0,1,-1\}$ (diagonalize it).