Question

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

Answer 1

Let $y=\frac{1+\alpha }{1-\alpha }$

$\Rightarrow \alpha =\frac{y-1}{y+1}$ the equation with roots $dfrac1+\alpha 1-\alpha ,\frac{1+\beta }{1-\beta },\frac{1+\gamma }{1-\gamma }$ is

$f\left(\frac{y-1}{y+1}\right)={\left(\frac{y-1}{y+1}\right)}^3+p\left(\frac{y-1}{y+1}\right)+r=0$

$\Rightarrow {(y-1)}^3+p(y-1){(y+1)}^2+r{(y+1)}^3=0$

$\Rightarrow y^3-1-3y^2+3y+p(y^2-1)(y+1)+r(y^3+3y^2+3y+1)=0$

$\Rightarrow y^3-1-3y^2+3y+p(y^3+2y^2+y-y^2-2y-1)+r(y^3+3y^2+3y+1)=0$

$\Rightarrow y^3(1+p+r)+y^2(-3+p+3r)+y(3-p+3r)-1-p+r=0$

$\therefore \frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }=\frac{3-p-3r}{1+p+r}$