One thing that jumps out at me is that this matrix has two eigenvalues. Find them: det $ (\begin{bmatrix} 2-\lambda & 3\\ 3 & 2-\lambda \end{bmatrix})$= $(2-\lambda)^{2}-9$=$-5-4\lambda+\lambda^{2}$. Factoring this you get $(\lambda+1)(\lambda-5)=0$, so your eigenvalues are -1 and 5. Now, find bases for their respective eigenspaces.
Basis $\xi_{-1}$ = ker$(A+I)$=ker $(\begin{bmatrix} 3 & 3\\\ 3 & 3 \end{bmatrix})$. Row-reducing, you get ker $(\begin{bmatrix} 1 & 1\\\ 0 & 0 \end{bmatrix})$, which is equal to the span of $\begin{bmatrix} -1\\ 1 \end{bmatrix}$. So a basis of $\xi_{-1}$ = $\begin{bmatrix} -1\\ 1 \end{bmatrix}$.
Same for $\xi_{5}$ -- ker $(\begin{bmatrix} -3 & 3\\\ 3 & -3 \end{bmatrix})$, row-reducing we get ker $(\begin{bmatrix} 1 & -1\\\ 0 & 0 \end{bmatrix})$, so ker = span $\begin{bmatrix} 1\\ 1 \end{bmatrix}$. So a basis of $\xi_{5}$ = $\begin{bmatrix} 1\\ 1 \end{bmatrix}$.
Now you know that $A$=$C^{3}$=$CCC$. So if $\vec{x}$ is an eigenvector of $A$, it is clearly an eigenvector of $C$, but perhaps with a different eigenvalue. You know that $A$ \begin{bmatrix} 1\\ 1 \end{bmatrix}$=$CCC$ \begin{bmatrix} 1\\ 1 \end{bmatrix}$=$\begin{bmatrix} 5\\ 5 \end{bmatrix}$. So, $\sqrt[3]{5}$ must be an eigenvalue of $C$.
Same goes for $A$\begin{bmatrix} -1\\ 1 \end{bmatrix}$=$CCC$\begin{bmatrix} -1\\ 1 \end{bmatrix}$=$\begin{bmatrix} 1\\ -1 \end{bmatrix}$. This would be possible only if -1 was an eigenvalue of $C$, since $(-1)^{3}$=$-1$.
So now you can construct a system of equations -- you know that $C$ is of some form $(\begin{bmatrix} a & b\\\ c & d \end{bmatrix})$, and based on the eigenvectors and eigenvalues of $C$ that we just found out, you can devise the following system:
$\left\{\begin{matrix} a & + & b & = & \sqrt[3]{5}\\ c & + & d & = & \sqrt[3]{5}\\ -a & + & b & = & 1\\ -c & + & d & = & -1 \end{matrix}\right.$
Now solve for the variables -- $b=a+1$, substitute into the other $a, b$ equation to get $2a+1=\sqrt[3]{5}$, so $a=\frac{\sqrt[3]{5}-1}{2}$. $b$, then, is equal to $\frac{\sqrt[3]{5}-1}{2}+1$, as obvious from the third equation. Same for $c$ and $d$ -- $c=d+1$, so substituting, we get $2d+1=\sqrt[3]{5}$, so $d=\frac{\sqrt[3]{5}-1}{2}$, and hence $c=\frac{\sqrt[3]{5}-1}{2}+1$.
This results into $C$=$ \begin{bmatrix} \frac{\sqrt[3]{5}-1}{2} & \frac{\sqrt[3]{5}-1}{2}+1 \\ \frac{\sqrt[3]{5}-1}{2}+1 & \frac{\sqrt[3]{5}-1}{2} \end{bmatrix}$.