Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3-x+1=0$, then $\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }=$

• A. 1
• B. 0
• C. 2
• D. -2

Answer 1

Answer: (A)

$\alpha ,\beta$ and $\gamma$ are the roots of $x^3-x+1=0\dots$ (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
${\left(\frac{y-1}{y+1}\right)}^3-\left(\frac{y-1}{y+1}\right)+1=0$
$\Rightarrow (y-1{)}^3-(y-1)(y+1{)}^2+(y+1{)}^3=0$
$\Rightarrow y^3-1+3y-3y^2-\left(y^2-1\right)$
$(y+1)+\left(y^3+1+3y+3y^2\right)=0$
$\Rightarrow y^3-1+3y-3y^2-y^3-y^2+y+1+y^3+1+3y+3y^2=0$
$\Rightarrow y^3-y^2+7y+1=0$
$\therefore$ Sum of roots $=\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }=\frac{-(-1)}{1}=1$