If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of the equation ${\mathrm{x}}^3–{\mathrm{ax}}^2+\mathrm{bx}–\mathrm{c}=0$ then $\sum {\mathrm{\alpha }}^2(\mathrm{\beta }+\mathrm{\gamma })=$

$\begin{array}{l}If\alpha ,\beta ,\gamma arerootsof{x}^3-a{x}^2+bx-c=0\\ Then\alpha +\beta +\gamma =a,ga\beta +\beta \gamma +\gamma \alpha =b,\alpha \beta \gamma =c\\ \therefore \sum {\alpha }^2(\beta +\gamma )={\alpha }^2(\beta +\gamma )+{\beta }^2(\gamma +\alpha )+{\gamma }^2(\alpha +\beta )\\ ={\alpha }^2(a-\alpha )+{\beta }^2(a-\beta )+{\gamma }^2(a-\gamma )\\ =a({\alpha }^2+{\beta }^2+{\gamma }^2)-({\alpha }^3+{\beta }^3+{\gamma }^3)--\left(1\right)\\ Now{\alpha }^2+{\beta }^2+{\gamma }^2={(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )\Rightarrow a^2-2b\\ and\\ {\alpha }^3-a{\alpha }^2+b\alpha -c=0\\ {\beta }^3-a{\beta }^2+b\beta -c=0\\ {\gamma }^3-a{\gamma }^3+b\gamma -c=0\\ \overline{({\alpha }^3+{\beta }^3+{\gamma }^3)-a({\alpha }^2+{\beta }^2+{\gamma }^2)+b(\alpha +\beta +\gamma )-3c=0}\end{array}$

$\begin{array}{l}\Rightarrow {\alpha }^3+{\beta }^3+{\gamma }^3=a({\alpha }^2+{\beta }^2+{\gamma }^2)-b(\alpha +\beta +\gamma )+3c\\ =a(a^2-2b)-b\left(a\right)+3c\\ =a^3-2ab-ba+3c\\ =a^3-3ab+3c\\ From\left(1\right)\\ \sum {\alpha }^2(\beta +\gamma )=a(a^2-2b)-a^3+3ab-3c\\ =a^3-2ab-a^3+3ab-3c\\ =ab-3c\end{array}$