Find a real matrix $B$ such that $B^3 = A$ 
Given $$A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}$$ find a real, invertible matrix $B$ such that $B^3 = A$


I think I am doing something wrong here, so let me describe my attempt:
1) So I started off with diagonalizing the matrix $A$ with finding the eigenvalues $\lambda_1 = -8$ and $\lambda_2 = 1$ and the corresponding eigenvectors $ \vec v_1 = \begin{bmatrix}1 & 1\\0 & 0\end{bmatrix} = x + y = 0 \Rightarrow -x = y \Rightarrow \begin{bmatrix}1\\-1\end{bmatrix}$ and $ \vec v_2 = \begin{bmatrix}1 & -\frac{1}{2}\\0 & 0\end{bmatrix} = x - \frac{1}{2}y = 0 \Rightarrow 2x = y \Rightarrow \begin{bmatrix}1\\2\end{bmatrix}$
2) With that being done I proceeded with computing $D = \begin{bmatrix}-8 & 0\\0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}$ and check everything with $D = PAP^{-1}$
3) Now I thought I will simple find a diagonal matrix $M = PBP^{-1}$ and $M^3 = D$ and the easiest solution I came up with was $M = \begin{bmatrix}\sqrt[3]{-8} & 0\\0& 1\end{bmatrix}$ so basically $M = D^{\frac{1}{3}}.$ So that $B = PMP^{-1}$. But now come the tricky part, if I compute $B$ it results in a complex matrix not a real. //It is real!
Have I perhaps overlooked something here or miscalculated the solution for $B$?

Edit:
As Cameron pointed out my calculator and I totally failed as it was in complex mode and computed one of the non-real cube roots instead of -2. So $M = \begin{bmatrix}-2 & 0\\0 & 1\end{bmatrix}$ and consequentially $B = \begin{bmatrix}-1 & 1\\2 & 0\end{bmatrix}$

The characteristic polynomial of $A$ is
\begin{align}
            p(\lambda)&= (-5-\lambda)(-2-\lambda)-18\\
                      &= \lambda^2+7\lambda-8 \\
                      &= (\lambda +8)(\lambda-1).
\end{align}
So $(A+8I)(A-I)=0$. That means
\begin{align}
              A(A+8I)&=(A+8I) \\
              A(A-I) &= -8(A-I) \\
          I &= \frac{1}{9}((A+8I)-(A-I)) \\
          A^{1/3}&=\frac{1}{9}((A+8I)+2(A-I)) \\
             &= \frac{1}{9}(3A+6I) = \frac{1}{3}(A+2I) \\
             &= \frac{1}{3}\begin{pmatrix}-3 & 3 \\ 6 & 0\end{pmatrix} 
            = \begin{pmatrix}-1 & 1 \\ 2 & 0 \end{pmatrix}.
\end{align}
