If α, β, γ are the roots of x^3-x+1=0, then (1+α)/(1-α)+(1+β)/(1-β)+(1+γ)/(1-γ)=

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MARKS Content Team · Accepted Answer

Correct answer: (A) 1. The given equation x^3-x+1=0 We know that If α,β,γ are the roots of the equation α+β+γ=(-b)/a,αβ+βγ+γα=c/a,αβγ=(-d)/a ∴α+β+γ=0, αβγ=-1, αβ+βγ+γα=-1 Given (1+α)/(1-α)+(1+β)/(1-β)+(1+γ)/(1-γ)=((1+α)(1−β)(1−γ)+(1−α)(1+β)(1−γ)+(1−α)(1−β)(1+γ) ​ )/((1-α)(1-β)(1-γ)) =(3−(α+β+γ)−(αβ+βγ+γα)+3(αβγ))/(1-(α+β+γ)−(αβ+βγ+γα)+(αβγ)) =(3-0-(-1)+3(-1))/(1-0-(-1)-1) =(3+1-3)/(1+1-1) =1

If $α, β, γ$ are the roots of $x^{3} - x + 1 = 0,$ then $\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ} =$

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Relevant Questions

If α, β, γ are the roots of x3-x+1=0, then 1+α1-α+1+β1-β+1+γ1-γ=

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Relevant Questions

If $α, β, γ$ are the roots of $x^{3} - x + 1 = 0,$ then $\frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} + \frac{1 + γ}{1 - γ} =$