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

$\alpha , \beta$ be the roots of ${\mathrm{x}}^2+\mathrm{px}+\mathrm{q}=0$

$\gamma , \delta$ be the roots of ${\mathrm{x}}^2+\mathrm{px}+\mathrm{r}=0$

$\begin{array}{l}\mathrm{\alpha }+\mathrm{\beta }=-\mathrm{p}\\ \mathrm{\alpha \beta }=\mathrm{q}\\ \mathrm{\gamma }+\mathrm{\delta }=-\mathrm{p}\\ \mathrm{\gamma \delta }=\mathrm{r}\end{array}$

Now, $(\mathrm{\alpha }-\mathrm{\gamma })(\mathrm{\alpha }-\mathrm{\delta })={\mathrm{\alpha }}^2-\mathrm{\alpha }(\mathrm{\gamma }+\mathrm{\delta })+\mathrm{\gamma \delta }$

$\begin{array}{l}={\mathrm{\alpha }}^2-\mathrm{\alpha }(\mathrm{\alpha }+\mathrm{\beta })+\mathrm{r}\\ =-\mathrm{\alpha \beta }+\mathrm{r}=-\mathrm{q}+\mathrm{r}\end{array}$

By symmetry $(\beta -\gamma )(\beta -\delta )=-q+r$

Hence, $\frac{(\alpha -\gamma )(\alpha -\delta )}{(\beta -\gamma )(\beta -\delta )}=1$