Let $A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right], B=\left[B_1, B_2, B_3\right]$, where $B_1, B_2, B_3$ are column matrics, and

$$ \mathrm{AB}_1=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right], \mathrm{AB}_2=\left[\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right], \quad \mathrm{AB}_3=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] $$

If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3+\beta^3$ is equal to ____________.

<p>$$\mathrm{A}=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] \quad \mathrm{B}=\left[\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3\right]$$</p> <p>$$\mathrm{B}_1=\left[\begin{array}{l} \mathrm{x}_1 \\ \mathrm{y}_1 \\ \mathrm{z}_1 \end{array}\right], \quad \mathrm{B}_2=\left[\begin{array}{l} \mathrm{x}_2 \\ \mathrm{y}_2 \\ \mathrm{z}_2 \end{array}\right], \quad \mathrm{B}_3=\left[\begin{array}{l} \mathrm{x}_3 \\ \mathrm{y}_3 \\ \mathrm{z}_3 \end{array}\right]$$</p> <p>$$\mathrm{AB}_1=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x}_1 \\ \mathrm{y}_1 \\ \mathrm{z}_1 \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]$$</p> <p>$$\begin{gathered} \mathrm{x}_1=1, \mathrm{y}_1=-1, \mathrm{z}_1=-1 \\ \mathrm{AB}_2=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x}_2 \\ \mathrm{y}_2 \\ \mathrm{z}_2 \end{array}\right]=\left[\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right] \\ \mathrm{x}_2=2, \mathrm{y}_2=1, \mathrm{z}_2=-2 \\ \mathrm{AB}_3=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x}_3 \\ \mathrm{y}_3 \\ \mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ \mathrm{x}_3=2, \mathrm{y}_3=0, \mathrm{z}_3=-1 \\ \mathrm{~B}=\left[\begin{array}{ccc} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{array}\right] \\ \alpha=|\mathrm{B}|=3 \\ \beta=1 \\ \alpha^3+\beta^3=27+1=28 \end{gathered}$$</p>