If α, β, γ are the roots of x3−x2−1=0 then the value of 1+α1−α+1+β1−β+1+γ1−γ is equal to

Σα=1,Σαβ=0,αβγ=1 ∑1+α1−α=−Σ−α+1−21−α=Σ21−α−1 =2Σ11−α−3Now, x3−x2−1=(x−α)(x−β)(x−γ)∴ log⁡(x−α)+log⁡(x−β)+log⁡(x−γ)=log⁡x3−x2−1Differentiating, we get1(x−α)+1(x−β)+1(x−γ)=3x2−2xx3−x2−1⇒11−α+11−β+11−γ=3−21−1−1=−1⇒∑ 1+α1−α=−5