Let $A \in \mathbb{C}^{3 \times 3}$ with $A^2(A^2 - 4 I) = 0$. What are possible minimal polynomials of $A^3$? Give examples of $A$ with their corresponding $A^3$ matrices.
At the moment I know that $f = A^2(A-2)(A+2)= 0$. Other then that I don't really know how to get the possible minimal polynomials of $A^3$. I was thinking of maybe using the property $a^3-b^3 = (a-b)(a^2 + ab + b^2)$, and thus multiplying the polynomial $f$ by $g = A(A^2 + 2A + 4)(A^2 - 2A + 4)$, so that I get $f\cdot g = A^3(A^3-8)(A^3+8)=0$ (since $f=0$).
Then I can continue looking for $A^3$ with possible eigenvalues 0, 8, -8. But I don't know if this is correct, nor do I know how I would find minimal polynomials for the matrix.
Does anybody know how I can properly find the minimal polynomials and is the method of multiplying it by $g$ a good idea?
HINT
You can greatly reduce your work by noting a couple of simple results.
Since $A^4=4A^2$ you know that $A^{12}=64A^6$. If $B=A^3$, then $B^4=64B^2$ and so $B$s minimal polynomial divides $x^4-64x^2$.
Also, since the matrices are $3\times 3$, the minimal polynomial is at most a cubic.
For each possibility check if you can find a suitable $A$ and $B$.
Addendum
In view of the comments below it might be helpful for me to add some general points about minimal polynomials.
The minimal polynomial $m(x)$ will divide any polynomial satisfied by the matrix. If you can find several polynomials the matrix has to satisfy then that will greatly reduce the possibilities.
Every root of the minimal polynomial is an eigenvalue. Again this could reduce the possibilities.
Every matrix satisfies the characteristic polynomial, $|A-xI|=0$. So $m(x)$ must divide this polynomial. (It might be considerably simpler than the characteristic polynomial.)