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

The given equation is $x^3-6x^2+11x+6=0$

$\alpha ,\beta ,\gamma$ are the roots of the equation

$\Rightarrow s_1=\alpha +\beta +\gamma =6,$

$s_2=\alpha \beta +\beta \gamma +\gamma \alpha =11,$

$s_3=\alpha \beta \gamma =-6$

$\therefore \sum {\alpha }^2\beta =\sum \alpha \sum \alpha \beta -3\left(\alpha \beta \gamma \right)$

                   $=6\left(11\right)+3\left(6\right)=84$