How to show that $ C := A^3-3A^2+2A = 0$? 
Let $$A = \begin{pmatrix} -1 & 1 & 2 \\ 0 & 2 & 0 \\ -1 & 1 & 2  \end{pmatrix}$$
Let $C:= A^3-3A^2+2A $.
Show that $C=0$.

I know that $A$ is diagonalisable with $\operatorname{spec}(A)=\{0,1,2\}$.
I have no clue how to approach that problem. Any advice?

Here is a way to obtain the value of $p(A)$, where $p(X)= X^3+3X^2+2X$, using Hamilton-Cayley:
The characteristic polynomial of $A$ is $\:\chi_A=X(X-1)(X-2)=X^3-3^2+2X$, and we have
$p(X)=\chi_A(X)-6X^2$, so
$$p(A)=\chi_A(A)-6A^2=-6A^2$$
so you have only one matrix multiplication to do.
