$$A=\begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}=PDP^{-1}$$
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 $$ B = PD^{1/3}P^{-1} = \begin{bmatrix}-1& 1\\2 & 0\end{bmatrix}$$
UPDATE:
After perforoming a row operation, it is easy for me to spot that
$$
(2, -1,-1,-1,-1,1,1)
$$
is a nontirivial solution as @user1551 commented. Thanks for that hint.
Disclaimer: this method might not be elegant enough as you wanted.
Solution. $\blacktriangleleft$ Add the $(-1)$ $\times$ 1st row to rows 2 thru 5, we have
$$
\det (\boldsymbol M) =: A =
\begin{vmatrix}
1&1&1&1&1&1&1\\
0&-2 &0 & 0 & 0 & -1 &-1\\
0& 0&-2 & 0 & 0 & -1 &-1\\
0& 0 &0 & -2 & 0 & -1 &-1\\
0& 0 &0 & 0 & -2 & -1 &-1\\
1& 0 &0 & 0 & 0 & -3 & 1\\
1& 0 &0 & 0 & 0 & 1 & -3
\end{vmatrix},
$$
Now add the $(-1)$ $\times$ 1st column to the 6th and 7th column:
$$
A = \begin{vmatrix}
1&1&1&1&1&0&0\\
0&-2 &0 & 0 & 0 & -1 &-1\\
0& 0&-2 & 0 & 0 & -1 &-1\\
0& 0 &0 & -2 & 0 & -1 &-1\\
0& 0 &0 & 0 & -2 & -1 &-1\\
1& 0 &0 & 0 & 0 & -4 & 0\\
1& 0 &0 & 0 & 0 & 0 & -4
\end{vmatrix}.
$$
Note that the right lower corner seems triangular, so we expand the determinant along the 1st column, then $A = A_1 - A_6 + A_7$ where the indices indicate the row.
$A_1$ is easy to calculate, since it is triangular, and we have $A_1 = 4^4 = 256$. For $A_6$:
$$
A_6 =
\begin{vmatrix}
1 & 1 & 1 & 1 & 0 &0\\
-2 &0 & 0 & 0 & -1 &-1\\
0&-2 & 0 & 0 & -1 &-1\\
0 &0 & -2 & 0 & -1 &-1\\
0 &0 & 0 & -2 & -1 &-1\\
0 &0 & 0 & 0 & 0 & -4
\end{vmatrix}
=
-4\begin{vmatrix}
1&1&1&1&0\\
-2 &0 & 0 & 0 & -1 \\
0&-2 & 0 & 0 & -1 \\
0 &0 & -2 & 0 & -1 \\
0 &0 & 0 & -2 & -1
\end{vmatrix}
$$
Add $1/2$ times rows of no. 2,3,4,5 to the 1st row:
$$
A_6 =
-4\begin{vmatrix}
0&0&0&0&-2\\
-2 &0 & 0 & 0 & -1 \\
0&-2 & 0 & 0 & -1 \\
0 &0 & -2 & 0 & -1 \\
0 &0 & 0 & -2 & -1
\end{vmatrix}
=
8\begin{vmatrix}
-2 &0 & 0 & 0\\
0&-2 & 0 & 0 \\
0 &0 & -2 & 0\\
0 &0 & 0 & -2
\end{vmatrix}
= 128.
$$
For $A_7$,
$$
A_7 =
\begin{vmatrix}
1 & 1 & 1 & 1 & 0 &0\\
-2 &0 & 0 & 0 & -1 &-1\\
0&-2 & 0 & 0 & -1 &-1\\
0 &0 & -2 & 0 & -1 &-1\\
0 &0 & 0 & -2 & -1 &-1\\
0 &0 & 0 & 0 & -4 & 0
\end{vmatrix}
= 4
\begin{vmatrix}
1 & 1 & 1 & 1 & 0\\
-2 &0 & 0 & 0 & -1\\
0&-2 & 0 & 0 & -1\\
0 &0 & -2 & 0 &-1\\
0 &0 & 0 & -2 &-1
\end{vmatrix}
= 4 \times (-32) = -128
$$
as calculated above. Hence $A = 256 - 128 + (-128) = 0$. $\blacktriangleright$
Best Answer
The Jordan form of $A$ is given by $$J = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ with $A = PJP^{-1}$ where $P = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 2\end{bmatrix}$.
First we can try to find $\log J$ blockwise. Searching for the logarithm of $\begin{bmatrix} -1 & 1 \\ 0 & -1\end{bmatrix}$ again as some Jordan block, if we set $f(x) = e^{-x}$ then $$\exp\left(\begin{bmatrix} i\pi & -1 \\ 0 & i\pi\end{bmatrix}\right) = f\left(\begin{bmatrix} -i\pi & 1 \\ 0 & -i\pi\end{bmatrix}\right) = \begin{bmatrix} f(-i\pi) & f'(-i\pi) \\ 0 & f(-i\pi)\end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 0 & -1\end{bmatrix}$$ so we can set (clearly it's not unique) $$\log J = \begin{bmatrix} i\pi & -1 & 0 \\ 0 & i\pi & 0 \\ 0 & 0 & 0\end{bmatrix}$$ and finally we get $$\log A = P(\log J) P^{-1} = \begin{bmatrix} 1 & -2 & -1 \\ 0 & 0 & 0 \\ 1 & -2 & -1\end{bmatrix}+i\pi \begin{bmatrix} 1 & 0 & 0 \\ -1 & 2 & 1 \\ 2 & -2 & -1\end{bmatrix}.$$