If x = 31, y = 32 and z = 33, then the value of $\left|\begin{array}{ccc}{\left({\mathrm{x}}^2+1\right)}^2& (\mathrm{xy}+1{)}^2& (\mathrm{xz}+1{)}^2\\ (\mathrm{xy}+1{)}^2& {\left({\mathrm{y}}^2+1\right)}^2& (\mathrm{yz}+1{)}^2\\ (\mathrm{xz}+1{)}^2& (\mathrm{yz}+1{)}^2& {\left({\mathrm{z}}^2+1\right)}^2\end{array}\right|$ is __________.

$\left|\begin{array}{ccc}{\left({\mathrm{x}}^2+1\right)}^2& (\mathrm{xy}+1{)}^2& (\mathrm{xz}+1{)}^2\\ (\mathrm{xy}+1{)}^2& {\left({\mathrm{y}}^2+1\right)}^2& (\mathrm{yz}+1{)}^2\\ (\mathrm{xz}+1{)}^2& (\mathrm{yz}+1{)}^2& {\left({\mathrm{z}}^2+1\right)}^2\end{array}\right|$

             $\begin{array}{c}=\left|\begin{array}{ccc}1& \mathrm{x}& {\mathrm{x}}^2\\ 1& \mathrm{y}& {\mathrm{y}}^2\\ 1& \mathrm{z}& {\mathrm{z}}^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 2\mathrm{x}& 2\mathrm{y}& 2\mathrm{z}\\ {\mathrm{x}}^2& {\mathrm{y}}^2& {\mathrm{z}}^2\end{array}\right|\\ =2(\mathrm{x}-\mathrm{y}{)}^2(\mathrm{y}-\mathrm{z}{)}^2(\mathrm{z}-\mathrm{x}{)}^2\end{array}$