Question

If $x_1,x_2,x_3$ be the roots of the equation $x^3-x+1=0$, then the value of $\left(\frac{1+x_1}{1-x_1}\right)\left(\frac{1+x_2}{1-x_2}\right)+\left(\frac{1+x_1}{1-x_1}\right)\left(\frac{1+x_3}{1-x_3}\right)+\left(\frac{1+x_2}{1-x_2}\right)\left(\frac{1+x_3}{1-x_3}\right)$ is

Answer 1

$x^3-x+1=0...\left(1\right)$
Let $y=\frac{1+x}{1-x}$
$\Rightarrow x=\frac{y-1}{y+1}$
From eqn(1),
${\left(\frac{y-1}{y+1}\right)}^3-\left(\frac{y-1}{y+1}\right)+1=0$
$\Rightarrow y^3-y^2+7y+1=0$
$\Rightarrow \sum \alpha \beta =7$
$\therefore \sum _{{}_{i\ne j}^{i=1,j=1}}^3\left(\frac{1+x_i}{1-x_i}\right)\left(\frac{1+x_j}{1-x_j}\right)=7$