If the sum of any two roots of the equation ${\mathrm{x}}^3+{\mathrm{px}}^2+\mathrm{qx}+\mathrm{r}=0$ is zero, then

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