If α, β and γ are the roots of the equation, x3-x-1=0 then, (1+α/1-α)+(1+β/1-β)+(1+γ/1-γ) has the value equal to:

Question

If α, β and &gamma; are the roots of the equation, x3-x-1=0 then, (1+α/1-α)+(1+β/1-β)+(1+&gamma;/1-&gamma;) has the value equal to: (A) zero (B) -1 (C) -7 (D) 1

Tardigrade Admin · Accepted Answer

y=(1+x/1-x) ⇒ x=(y-1/y+1) ⇒((y-1/y+1))3-(y-1/y+1)-1=0 ⇒ y3+7 y2-y+1=0 ⇒ ∑ (1+α/1-α)=-7

Q. If $α, β$ and $γ$ are the roots of the equation, $x^{3} - x - 1 = 0$ then, $\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ}$ has the value equal to:

zero

-1

-7

1

$y = \frac{1 + x}{1 - x} \Rightarrow x = \frac{y - 1}{y + 1} \Rightarrow (\frac{y - 1}{y + 1})^{3} - \frac{y - 1}{y + 1} - 1 = 0$
$\Rightarrow y^{3} + 7 y^{2} - y + 1 = 0 \Rightarrow \sum \frac{1 + α}{1 - α} = - 7$

Q. If α,β and γ are the roots of the equation, x3−x−1=0 then, 1−α1+α​+1−β1+β​+1−γ1+γ​ has the value equal to:

zero

-1

-7

1

y=1−x1+x​⇒x=y+1y−1​⇒(y+1y−1​)3−y+1y−1​−1=0 ⇒y3+7y2−y+1=0⇒∑1−α1+α​=−7

Q. If $α, β$ and $γ$ are the roots of the equation, $x^{3} - x - 1 = 0$ then, $\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ}$ has the value equal to:

$y = \frac{1 + x}{1 - x} \Rightarrow x = \frac{y - 1}{y + 1} \Rightarrow (\frac{y - 1}{y + 1})^{3} - \frac{y - 1}{y + 1} - 1 = 0$
$\Rightarrow y^{3} + 7 y^{2} - y + 1 = 0 \Rightarrow \sum \frac{1 + α}{1 - α} = - 7$