When looking for the eigenvalue of A, doing the determinant I get terms in cube which I don't know how to factorise. I know how to do them when I can expand my DET along a row/column with 2 zeroes so that I only have a cube to factorise but here I am stuck... How would you proceed?
For example with $\;A\;$ :
$$\det(tI-A)=\begin{vmatrix} t-2&0&-8\\ -1&t-4&4\\ -1&-2&t+3\end{vmatrix}=(t-2)(t^2-t-12)-16+8(t-2)-8(t-4)=$$$${}$$
$$=t^3-3t^2-10t+24=(t+3)(t^2-6t+8)=(t+3)(t-2)(t-4)$$
and thus you have three distinct eigenvalues for a $\;3\times3\;$ matrix, so it is...