$\begin{aligned} & A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] \quad B=\left[B_1, B_2, B_3\right] \\ & B_1=\left[\begin{array}{l}x_1 \\ y_1 \\ z_1\end{array}\right], \quad B_2=\left[\begin{array}{l}x_2 \\ y_2 \\ z_2\end{array}\right], \quad B_3=\left[\begin{array}{l}x_3 \\ y_3 \\ z_3\end{array}\right] \\ & A_1=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ y_1 \\ z_1\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\end{aligned}$

$\mathrm{x}_1=1, \mathrm{y}_1=-1, \mathrm{z}_1=-1$

$\begin{aligned} & \mathrm{AB}_2=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x}_2 \\ \mathrm{y}_2 \\ \mathrm{z}_2\end{array}\right]=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right] \\ & \mathrm{x}_2=2, \mathrm{y}_2=1, \mathrm{z}_2=-2 \\ & \mathrm{AB}_3=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x}_3 \\ \mathrm{y}_3 \\ \mathrm{z}_3\end{array}\right]=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right] \\ & \mathrm{x}_3=2, \mathrm{y}_3=0, \mathrm{z}_3=-1 \\ & \mathrm{~B}=\left[\begin{array}{ccc}1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1\end{array}\right] \\ & \alpha=|\mathrm{B}|=3 \\ & \beta=1 \\ & \alpha^3+\beta^3=27+1=28\end{aligned}$