If $\tan \mathrm{\alpha },\tan \mathrm{\beta },\tan \mathrm{\gamma }$ are the roots of the equation ${\mathrm{x}}^3-{\mathrm{px}}^2-\mathrm{r}=0$ then the value of $\left(1+{\tan }^{2 }\mathrm{\alpha }\right)\left(1+{\tan }^2 \mathrm{\beta }\right)\left(1+{\tan }^2\mathrm{\gamma }\right)=$

$\sum \tan \mathrm{\alpha }=\mathrm{p},\sum \tan \mathrm{\alpha tan} \mathrm{\beta }=0,\tan \mathrm{\alpha tan} \mathrm{\beta tan} \mathrm{\gamma }=\mathrm{r}.$
Required value value
$\begin{array}{l}=1+{\left(\sum \tan \mathrm{\alpha }\right)}^2-2\sum \tan \mathrm{\alpha tan}\mathrm{\beta }+{\left(\sum \tan \mathrm{\alpha tan}\mathrm{\beta }\right)}^2-\\ 2\tan \mathrm{\alpha tan} \mathrm{\beta tan} \mathrm{\gamma }\sum \tan \mathrm{\alpha }+(\tan \mathrm{\alpha tan} \mathrm{\beta tan} \mathrm{\gamma }{)}^2\\ =1+{\mathrm{p}}^2-2\mathrm{pr}+{\mathrm{r}}^2=1+(\mathrm{p}-\mathrm{r}{)}^2\end{array}$