Find a real matrix $B$ such that $B^3 = A$

$$A=\begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}=PDP^{-1}$$

Where $$P=\begin{bmatrix}1& 1\\-1 & 2\end{bmatrix}$$is the matrix of eigenvectors

and $$D=\begin{bmatrix}-8& 0\\0 & 1\end{bmatrix}$$ is the matrix of eigenvalues.

Thus $$ B = PD^{1/3}P^{-1} = \begin{bmatrix}-1& 1\\2 & 0\end{bmatrix}$$

How in the world would the product of three real matrices turn out to have non-real entries? Something must have gone wrong. Try rewriting $\sqrt[3]{-8}=-2,$ and see if that makes a difference. It may be that your calculator chose one of the two non-real cube roots of $-8,$ instead, or perhaps you accidentally entered some even root of $-8$.

For a $2\times 2$ matrix $\{A\}$ with distinct eigenvalues $\{\lambda_1, \lambda_2\}$, any function $\{f(A)\}$ can be evaluated as a linear polynomial, whose coefficients are determined solely by the eigenvalues $$\eqalign{ c_1 &= \tfrac{f(\lambda_1)-f(\lambda_2)}{\lambda_1-\lambda_2}\cr c_0 &= f(\lambda_1) - c_1\lambda_1 \cr f(A) &= c_1A + c_0I \cr\cr }$$ For the current problem, $B=f(A)=A^{1/3}$ and $(\lambda_1,\lambda_2)=(-8,1)$, therefore $$\eqalign{ f(A) &= \tfrac{1}{3}A + \tfrac{2}{3}I\cr }$$