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

The correct option is D

$-5$

Roots of $x^3-x^2-1=0$ are $\alpha ,\beta ,\gamma$.

Let $y=\frac{1+x}{1-x}\Rightarrow x=\frac{y-1}{y+1}$.

Then $\frac{(y-1{)}^3}{(y+1{)}^3}-{\left(\frac{y-1}{y+1}\right)}^2-1=0$

i.e., $(y-1{)}^3-(y-1{)}^2(y+1)-(y+1{)}^3=0$

i.e., $y^3+5y^2-y+1=0$ has roots

$\frac{1+\alpha }{1-\alpha },\frac{1+\beta }{1-\beta },\frac{1+\gamma }{1-\gamma }$,
whose sum $=-5$