Diagonalization of matrix and eigenvalues of $A+A^2+A^3$ Let $$A=
        \begin{pmatrix}
        6 & 4 & 1 \\
        -6 & -1 & 3 \\
        8 & 8 & 4 \\
        \end{pmatrix}
$$
Find a non-sigular matrix P and a diagonal matrix D such that $A+A^2+A^3=PDP^{-1}$
No idea what to do

You do not have to calculate $A+A^2+A^3$.
Suppose that you know how to diagonalize the matrix $A$, i.e., you can find an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$.
(I will leave the computations to you. You can have a look at other posts tagged diagonalization or on Wikipedia. I guess you can find there something to get you started. Then you can check your result on WolframAlpha or using some other tools.) 
If you already have $A=PDP^{-1}$, then it is easy to see that
$$
\begin{align*}
A^2&=(PDP^{-1})(PDP^{-1}) = PD^2P^{-1}\\
A^3&=(PD^2P^{-1})PDP^{-1} = PD^3P^{-1}
\end{align*}
$$
and we get
$$A+A^2+A^3=PDP^{-1}+PD^2P^{-1}+PD^3P^{-1} = P(D+D^2+D^3)P^{-1}.$$
Clearly, the matrix $D+D^2+D^3$ is a diagonal matrix. If $d_1$, $d_2$, $d_3$ are the diagonal elements of the matrix the, then
$D+D^2+D^3$ has
$$d'_i=d_i+d_i^2+d_i^3$$
on the diagonal.
