Roots and calcul

**dhiab** · September 5th 2009, 11:48 PM

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

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

**red_dog** · September 6th 2009, 12:27 AM

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 $y_1+y_2+y_3=-7$

**pacman** · September 6th 2009, 04:23 AM

red_dog bashed the olympiad caliber problem again, whew!

**I4talent** · September 7th 2009, 10:33 PM

$x^3-x-1 = 0$

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

$\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 $

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

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

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

**dhiab** · September 7th 2009, 11:07 PM

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

**pacman** · September 8th 2009, 05:08 AM

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

**pacman** · September 8th 2009, 05:12 AM

thanks

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