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Q.
If $\alpha, \beta, \gamma$ are the roots of $x^3-x+1=0$, then
$\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=$
TS EAMCET 2021
Solution:
$\alpha, \beta$ and $\gamma$ are the roots of $x^3-x+1=0 \ldots$ (i)
We have to find a polynomial whose roots are
$\frac{1+\alpha}{1-\alpha}, \frac{1+\beta}{1-\beta}, \frac{1+\gamma}{1-\gamma}$
Let $y=\frac{1+x}{1-x} \Rightarrow x=\frac{y-1}{y+1}$
Substituting $x=\frac{y-1}{y+1}$ in Eq. (i), we have
$ \left(\frac{y-1}{y+1}\right)^3-\left(\frac{y-1}{y+1}\right)+1=0 $
$ \Rightarrow (y-1)^3-(y-1)(y+1)^2+(y+1)^3=0$
$ \Rightarrow y^3-1+3 y-3 y^2-\left(y^2-1\right)$
$ (y+1)+\left(y^3+1+3 y+3 y^2\right)=0$
$ \Rightarrow y^3-1+3 y-3 y^2-y^3-y^2+y+1 +y^3+1+3 y+3 y^2=0$
$\Rightarrow y^3-y^2+7 y+1=0$
$\therefore$ Sum of roots $=\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=\frac{-(-1)}{1}=1$