Given a matrix $A$ find a matrix $C$ such that $C^3$=$A$

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}$.

This would be easy for a diagonal matrix, because $\begin{bmatrix}a & 0\\ 0 & b\end{bmatrix}^3=\begin{bmatrix}a^3 & 0\\ 0 & b^3\end{bmatrix}$, which means that you could just take the cube root of each diagonal entry to solve the problem. While $A$ is not diagonal, it is symmetric and therefore diagonalizable. If you're comfortable with diagonalizing, find $S$ such that $SAS^{-1}=\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}$. You know how to find a matrix whose cube gives the right hand side. Then notice how conjugation behaves with cubing: $(S^{-1}C'S)^3=S^{-1}C'^3S$. Therefore, you can take $C=S^{-1}\begin{bmatrix}\sqrt[3]{a} & 0 \\ 0 & \sqrt[3]{b} \end{bmatrix}S$.

(I was going to write more involving eigenvectors, but then other answers were posted covering this.)

An alternative approach using polynomial interpolation will work for all diagonalizable matrices having eigenvalues $-1$ and $5$, and does not require finding eigenvectors. For more on this and generalizations, see Chapter 1 of Higham's Functions of matrices.

In this case, the Lagrange interpolating polynomial of the cube root function on the spectrum $\{5,-1\}$ of $A$ is $$p(t)=\sqrt[3]{5}\cdot\frac{t+1}{5+1}+\sqrt[3]{-1}\cdot\frac{t-5}{-1-5}=\frac{\sqrt[3]{5}+1}{6}\cdot t +\frac{\sqrt[3]{5}-5}{6},$$ so that a matrix cube root for $A$ can be obtained as $$p(A)=\frac{\sqrt[3]{5}+1}{6}\cdot A +\frac{\sqrt[3]{5}-5}{6}\cdot I=\frac{1}{2}\begin{bmatrix}\sqrt[3]5-1&\sqrt[3]5+1\\\sqrt[3]5+1&\sqrt[3]5-1\end{bmatrix}.$$

Remember that:

Matrices act on vectors.

Here the linearly independent vectors $u=\begin{pmatrix} 1\\ 1\end{pmatrix}$ and $v=\begin{pmatrix} 1\\ -1\end{pmatrix}$ are such that $Au=5u$ and $Av=-v$.

Hence a suitable $C=\begin{pmatrix} a & b\\ c& d\end{pmatrix}$ could be defined by the conditions that $Cu=5^{1/3}u$ and $Cv=-v$. This gives you a linear system of two equations for $(a,b)$ and another linear system of two equations for $(c,d)$, and the matrix $C$ follows.