If $\mathrm{x}+\mathrm{y}+\mathrm{z}=12\text{ and }{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=96$ and $\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=36$, then the value of ${\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3$ is_______.

$\begin{array}{l}(\mathrm{x}+\mathrm{y}+\mathrm{z}{)}^2=144\text{ (given) }\\ \Rightarrow \sum {\mathrm{x}}^2+2\sum \mathrm{xy}=144\\ \Rightarrow 96+2\sum \mathrm{xy}=144\Rightarrow \sum \mathrm{xy}=24\end{array}$
Again $\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=36\Rightarrow \mathrm{xyz}=\frac{24}{36}=\frac{2}{3}$
$\begin{array}{l}\Rightarrow \sum {\mathrm{x}}^3-2=\left(12\right)(96-24)=\left(12\right)\left(72\right)=864\\ \Rightarrow \sum {\mathrm{x}}^3=866\end{array}$