If $\alpha ,\beta ,\gamma$ be the roots of x³ + px² + qx + r = 0, then find a cubic equation whose roots are $\alpha (\beta +\gamma ),\beta (\gamma +\alpha ),\gamma (\alpha +\beta )$

Let 

$\begin{array}{l}y=\alpha \beta +\alpha \gamma +\beta \gamma -\beta \gamma \\ y=q-\beta \gamma \\ y=q-\frac{\alpha \beta \gamma }{\alpha }=q+\frac{r}{\alpha }\\ \alpha =\frac{r}{y-q}\end{array}$

Since $\alpha$ is a root of the given equation, so

$\begin{array}{l}{\alpha }^3+p{\alpha }^2+q\alpha +r=0\\ {\left(\frac{r}{y-q}\right)}^3+p{\left(\frac{r}{y-q}\right)}^2+q\left(\frac{r}{y-q}\right)+r=0\end{array}$

Hence, the required equation is

$r^3+pr(x-q)+q(x-q{)}^2+r(x-q{)}^3=0$