If - - - are the roots of x3-x2-1 - 0- then the value of 1-1-1-1-1-1- is equal to-

Roots of x^3 x^2 1=0 are α, β, γ. Let y = 1+x1 x⇒ x nbsp;= y 1y+1. Then (y 1)^3(y+1)^3 ( y 1y+1 )^2 1 = 0 i.e., (y 1)^3 (y 1)^2(y+1) (y+1)^3 = 0 i. ...

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Byju's Answer

Standard XII

Mathematics

Cubic Polynomial

If α ,β ,γ ar...

Question

If $α, β, γ$ are the roots of $x^{3} - x^{2} - 1 = 0$ , then the value of $\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ}$ is equal to:

$- 7$

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$- 6$

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$- 5$

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Solution

The correct option is D
$- 5$

Roots of $x^{3} - x^{2} - 1 = 0$ are $α, β, γ$ .

Let $y = \frac{1 + x}{1 - x} \Rightarrow x = \frac{y - 1}{y + 1}$ .

Then $\frac{(y - 1)^{3}}{(y + 1)^{3}} - {(\frac{y - 1}{y + 1})}^{2} - 1 = 0$

i.e., $(y - 1)^{3} - (y - 1)^{2} (y + 1) - (y + 1)^{3} = 0$

i.e., $y^{3} + 5 y^{2} - y + 1 = 0$ has roots

$\frac{1 + α}{1 - α}, \frac{1 + β}{1 - β}, \frac{1 + γ}{1 - γ}$ ,
whose sum $= - 5$

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If $α, β, γ$ are the roots of $x^{3} - x^{2} - 1 = 0$ , then the value of $\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ}$ is equal to: