If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of ${\mathrm{x}}^3-{\mathrm{x}}^2-1=0$ then the value of $\frac{1+\mathrm{\alpha }}{1-\mathrm{\alpha }}+\frac{1+\mathrm{\beta }}{1-\mathrm{\beta }}+\frac{1+\mathrm{\gamma }}{1-\mathrm{\gamma }}$ is equal to 

$\begin{array}{l}\mathrm{\Sigma \alpha }=1,\mathrm{\Sigma \alpha \beta }=0,\mathrm{\alpha \beta \gamma }=1\\ \begin{array}{rl}\sum \frac{1+\mathrm{\alpha }}{1-\mathrm{\alpha }}& =-\mathrm{\Sigma }\frac{-\mathrm{\alpha }+1-2}{1-\mathrm{\alpha }}=\mathrm{\Sigma }\left(\frac{2}{1-\mathrm{\alpha }}-1\right)\\ & =2\mathrm{\Sigma }\frac{1}{1-\mathrm{\alpha }}-3\end{array}\end{array}$

Now, ${\mathrm{x}}^3-{\mathrm{x}}^2-1=(\mathrm{x}-\mathrm{\alpha })(\mathrm{x}-\mathrm{\beta })(\mathrm{x}-\mathrm{\gamma })$

$\therefore \log (\mathrm{x}-\mathrm{\alpha })+\log (\mathrm{x}-\mathrm{\beta })+\log (\mathrm{x}-\mathrm{\gamma })=\log \left({\mathrm{x}}^3-{\mathrm{x}}^2-1\right)$

Differentiating, we get

$\begin{array}{l}\frac{1}{(\mathrm{x}-\mathrm{\alpha })}+\frac{1}{(\mathrm{x}-\mathrm{\beta })}+\frac{1}{(\mathrm{x}-\mathrm{\gamma })}=\frac{3{\mathrm{x}}^2-2\mathrm{x}}{{\mathrm{x}}^3-{\mathrm{x}}^2-1}\\ \Rightarrow \frac{1}{1-\mathrm{\alpha }}+\frac{1}{1-\mathrm{\beta }}+\frac{1}{1-\mathrm{\gamma }}=\frac{3-2}{1-1-1}=-1\\ \Rightarrow \sum _{ }\frac{1+\mathrm{\alpha }}{1-\mathrm{\alpha }}=-5\end{array}$