How do I solve this matrix equation $X^3=\begin{pmatrix}19&30\\ -45&-71\end{pmatrix}$? I have to solve for the $2\times 2$ matrix $X$ the following equation

$$X^3=\begin{pmatrix}19&amp;30\\ -45&amp;-71\end{pmatrix}$$

To be honest I have no idea how to start.

I know how to solve equations like 
$$A \cdot X=C =&gt; X=A^{-1} \cdot C$$

$$X \cdot B=C =&gt; X=C \cdot B^{-1}$$

$$A \cdot X \cdot B=C =&gt; X=A^{-1} \cdot C \cdot B^{-1}$$

but I don't know how I could apply it in here
The given matrix $m$ has determinant $1$ and trace = $-52$. Assuming $X$ is a real matrix, $det(X)=1$. Let $t= tr(X)$. Now $X$ satisfies the equation $x^2-tx+1=0$.
From this it is easy to see that $m=X^3=(t^2-1)X-t*I_2$ , where $I_2$ is the identity matrix. So $(t^2-1)X=m+t*I_2$. Taking trace of both sides, we see that $t$ is a root of the cubic equation $t^3-3t=-52$ whose only real solution is $t=-4$. Therefore $15X=m-4I_2$. This gives 

$$X=\begin{pmatrix}1&amp;2\\ -3&amp;-5\end{pmatrix}$$
I'm not providing here a method to provide the result directly, but I would decompose the resulting matrix into simple geometrical transformation : dilatation, rotation ..., then I would divide the dilatation coefficient by 3, the angle by 3 etc ...

That would be my approach.