If $\mathrm{\alpha },\mathrm{\beta }$ are the roots of ${\mathrm{x}}^2+\mathrm{px}-\mathrm{q}=0$ and $\mathrm{\gamma },\mathrm{\delta }$ are the roots of ${\mathrm{x}}^2+\mathrm{px}+\mathrm{r}=0$ then $\left(\mathrm{\alpha }-\mathrm{\gamma }\right)\left(\mathrm{\beta }-\mathrm{\gamma }\right)\left(\mathrm{\alpha }-\mathrm{\delta }\right)\left(\mathrm{\beta }-\mathrm{\delta }\right)=$

$\begin{array}{l}If\alpha ,\beta arerootsof{x}^2+px+q=0\\ \alpha +\beta =-p,\alpha \beta =-q\\ and{\alpha }^2+p\alpha -q=0\Rightarrow {\alpha }^2+p\alpha =q\\ {\beta }^2+p\beta -q=0\Rightarrow {\beta }^2+p\beta =q\\ If\gamma ,\delta arerootsof{x}^2+px+r=0\\ \gamma +\delta =-p,\gamma \delta =r\\ (\alpha -\gamma )(\beta -\gamma )(\alpha -\delta )(\beta -\delta )=({\alpha }^2-(\gamma +\delta )\alpha +\gamma \delta )({\beta }^2-(\gamma +\delta )\beta +\gamma \delta )\\ =({\alpha }^2+p\alpha +\gamma )({\beta }^2+p\beta +\gamma )\\ =(q+r)(q+r)\\ ={(q+r)}^2\end{array}$