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 $B = \begin{bmatrix}-1 & 1\\2 & 0\end{bmatrix}$
 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 $$ B = PD^{1/3}P^{-1} = \begin{bmatrix}-1& 1\\2 & 0\end{bmatrix}$$
A: 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 $-8$.
A: 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
$$\eqalign{
 f(A) &= \tfrac{1}{3}A + \tfrac{2}{3}I\cr
}$$
A: 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}
A: Note: My goal here was to work the problem using triangular matrices. 
Let
$
A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}.
$
The eigenvalues for the matrix $A$ are $1$ and $-8$.
Set
$
\quad R = \begin{bmatrix}-2 & 1 \\ 0 & 1 \end{bmatrix} \quad \text{and} \quad S = \begin{bmatrix}-8 & 3 \\ 0 & 1 \end{bmatrix}
$
Observe that
$\tag 1 R^3 = S$
Set
$
\quad T = \begin{bmatrix}1 & 0 \\ 1 & 1 \end{bmatrix}
$
Observe that
$\tag 2 T^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$
and that
$\tag 3 T\,A\,T^{-1} = S$
Set
$\quad B = T^{-1} \, R \, T$
Using algebra it is easy to see that $B^3 = A$. 
Calculating
$
\quad B = \begin{bmatrix}-1 & 1 \\ 2 & 0 \end{bmatrix}
$
A: Standard computer methods (e.g., Mathematica Solve) gives the answer directly:
myB = {{b11, b12}, {b21, b22}}; 
myB /. Solve[myB.myB.myB == {{-5, 3}, {6, -2}}, {b11, b12, b21, b22}][[1]]

(*
{{-1, 1}, {2, 0}}
*)
$${\bf B} = {-1, 1 \choose 2 , 0}$$.
