1. ## Roots and calcul

$\alpha ,\beta ,\gamma$ roots of equation :

$x^3 - x - 1 = 0$
Calculate :
$\frac{{1 + \alpha }}{{1 - \alpha }} + \frac{{1 + \beta }}{{1 - \beta }} + \frac{{1 + \gamma }}{{1 - \gamma }}$

2. Let $y_1=\frac{1+\alpha}{1-\alpha}, \ y_2=\frac{1+\beta}{1-\beta}, \ y_3=\frac{1+\gamma}{1-\gamma}$.

We'll find the equation whose roots are $y_1, y_2, y_3$

For this let $y=\frac{1+x}{1-x}\Rightarrow x=\frac{y-1}{y+1}$

Replace x in the equation:

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

Then $y_1+y_2+y_3=-7$

3. red_dog bashed the olympiad caliber problem again, whew!

4. $x^3-x-1 = 0$

$(x-\alpha)(x-\beta)(x-\gamma) = 0$

$\left(\frac{1+x}{1-x}-\frac{1-\alpha}{1+\alpha}\right) \left(\frac{1+x}{1-x}-\frac{1-\beta}{1+\beta}\right) \left(\frac{1+x}{1-x}-\frac{1-\gamma}{1+\gamma}\right) = 0
$

$\left(y-\frac{1-\alpha}{1+\alpha}\right) \left(y-\frac{1-\beta}{1+\beta}\right) \left(y-\frac{1-\gamma}{1+\gamma}\right) = 0
$

$
\left (\frac{y-1}{1+y}-\alpha\right) \left (\frac{y-1}{1+y}-\beta\right) \left (\frac{y-1}{1+y}-\gamma\right) = 0
$

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

$
\frac{(y-1)(y-1)(y-1)}{(y+1)(y+1)(y+1)}-\left(\frac{y-1}{1+y}\right)-1 = 0$

$(1+y)(1+y)(1+y) \frac{(y-1)(y-1)(y-1)}{(y+1)(y+1)(y+1)}-$ $(1+y)(1+y)(1+y)\left(\frac{y-1}{1+y}\right)-(1+y)(1+y)(1+y) = 0$

$(y-1)(y-1)(y-1)-(1+y)(1+y)(y-1)-(1+y)(1+y)(1+y) = 0$

$-1+y^3-3y^2+3y-(-1-y+y^2+y^3)-(1+3y+3y^2+y^3) = 0$

$-1+y^3-3y^2+3y+1+y-y^2-y^3-1-3y-3y^2-y^3 = 0$

$y^3-y^3-y^3-3y^2-y^2-3y^2+3y+y-3y+1-1-1 = 0
$

$-y^3-7y^2+y-1 = 0$

$y^3+7y^2-y+1 = 0$

$\therefore\frac{{1 + \alpha }}{{1 - \alpha }} + \frac{{1 + \beta }}{{1 - \beta }} + \frac{{1 + \gamma }}{{1 - \gamma }} = \frac{-7}{1} = \boxed{-7}$

5. Originally Posted by red_dog
Let $y_1=\frac{1+\alpha}{1-\alpha}, \ y_2=\frac{1+\beta}{1-\beta}, \ y_3=\frac{1+\gamma}{1-\gamma}$.

We'll find the equation whose roots are $y_1, y_2, y_3$

For this let $y=\frac{1+x}{1-x}\Rightarrow x=\frac{y-1}{y+1}$

Replace x in the equation:

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

Then $y_1+y_2+y_3=-7$
Hello: Are you the details for y1 + y 2+ y 3 =-7 Thanks

6. i told you dhiab, red_dog's method is great. It was even beautiful than the official solution, or that of Andreescu and Feng.

Neat , and concise

7. thanks