If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are the roots of the equation${\mathrm{x}}^3–6{\mathrm{x}}^2+11\mathrm{x}+6=0$ then $\sum {\mathrm{\alpha }}^2\mathrm{\beta }=$ 

$\begin{array}{l}If\alpha ,\beta ,\gamma arerootsof{x}^3-6x^2+11x+6=0\\ Then\alpha +\beta +\gamma =6,\alpha \beta +\beta \gamma +\gamma \alpha =11,\alpha \beta \gamma =-6\\ \sum {\alpha }^2\beta ={\alpha }^2\beta +{\beta }^2\gamma +{\gamma }^2\alpha +{\gamma }^2\beta +{\alpha }^2\gamma +{\beta }^2\alpha \\ =\alpha \beta (\alpha +\beta )+\beta \gamma (\beta +\gamma )+\alpha \gamma (\alpha +\gamma )\\ =\alpha \beta (6-\gamma )+\beta \gamma (6-\alpha )+\alpha \gamma (6-\beta )\\ =6(\alpha \beta +\beta \gamma +\gamma \alpha )\\ =66+18\\ =84\end{array}$