Question

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

• A. zero
• B. -1
• C. -7
• D. 1

Answer 1

Answer: (C)

$y=\frac{1+x}{1-x} \Rightarrow x=\frac{y-1}{y+1} \Rightarrow\left(\frac{y-1}{y+1}\right)^3-\frac{y-1}{y+1}-1=0$
$\Rightarrow y^3+7 y^2-y+1=0 \Rightarrow \sum \frac{1+\alpha}{1-\alpha}=-7$