If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are the roots of the equation ${\mathrm{x}}^3 + {\mathrm{px}}^2 + \mathrm{qx} + \mathrm{r} = 0$ then $\sum {\mathrm{\alpha }}^2\left(\mathrm{\beta }+\mathrm{\gamma }\right)=$

$\begin{array}{l}If\alpha ,\beta ,\gamma areroots{x}^3+p{x}^2+qx+r=0\\ Then\alpha +\beta +\gamma =-p,\alpha \beta +\beta \gamma +\gamma \alpha =q,\alpha \beta \gamma =-r\\ \sum {\alpha }^2(\beta +\gamma )={\alpha }^2(\beta +\gamma )+{\beta }^2(\alpha +\gamma )+{\gamma }^2(\alpha +\beta )\\ ={\alpha }^2(-p-\alpha )+{\beta }^2(-p-\beta )+{\gamma }^2(-p-\gamma )\\ =-{\alpha }^{2}p-{\alpha }^3-p{\beta }^2-{\beta }^3-p{\gamma }^2-{\gamma }^3\\ =-p({\alpha }^2+{\beta }^2+{\gamma }^2)-({\alpha }^3+{\beta }^3+{\gamma }^3)\\ =-p({(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha ))-{(\alpha +\beta +\gamma )}^3+3\alpha \beta \gamma (\alpha +\beta +\gamma )\\ =-p(p^2-2q)-{(-p)}^3+3(-r)(-p)\\ =-p^3+2pq+p^3+3pr\because 2pq+p^3\end{array}$