If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of ${\mathrm{x}}^3–{\mathrm{px}}^2+\mathrm{qx}–\mathrm{r}=0$
then ${\mathrm{\alpha }}^3+{\mathrm{\beta }}^3+{\mathrm{\gamma }}^3=$
 

$\begin{array}{l}If\alpha ,\beta ,\gamma arerootsof{x}^3-p{x}^2+qx-r=0then\\ \Rightarrow {\alpha }^3-p{\alpha }^2+q\alpha -r=0\\ \Rightarrow {\beta }^3-p{\beta }^3+q\beta -r=0\\ \Rightarrow {\gamma }^3-p{\gamma }^2+q\gamma -r=0\\ \overline{\Rightarrow {\alpha }^3+{\beta }^3+{\gamma }^3-p({\alpha }^2+{\beta }^2+{\gamma }^2)+q(\alpha +\beta +\gamma )-3r=0}\\ \therefore {\alpha }^3+{\beta }^3+{\gamma }^3=p({\alpha }^2+{\beta }^2+{\gamma }^2)-q(\alpha +\beta +\gamma )+3r\\ =p({(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha ))-q(\alpha +\beta +\gamma )+3r\\ =p(p^2-2q)-pq+3r\\ =p^3-2pq-pq+3r\\ =p^3-3pq+3r\end{array}$