If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ Where $M$ is the minimal polynomial
$x^3-x=x(x^2-1)=x(x-i)(x+i)$ so $A$ is diagonalizable 
$A=P^{-1}\begin{pmatrix} 
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}P$ 
$A^2=P^{-1}\begin{pmatrix} 
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}^2P=P^{-1}\begin{pmatrix} 
0 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}^2P$ 
So $P_{A^2}(x)=x(x-1)^2$ but $A^2(A^2-1)=0$ so $M_{A^2}(x)=x(x-1)=x^2-x$
Is there another way to conclude it? 

First you can plug in $A^2$ to see whether it satisfies the polynomial. Indeed:
$$(A^2)^2 - A^2 = A^4 - A^2 = A A^3 - A^2 = AA - A^2=0$$
Now check that $A^2$ doesn't satisfy the linear factors $x$ and $x-1$.
