Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3-x-1=0$, then the value of $\Pi \left(\frac{1+\alpha }{1-\alpha }\right)$ is equal to -

• A. $-7$
• B. $-5$
• C. $-3$
• D. $-1$

Answer 1

Answer: (D)

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