If α,β,γ are roots of x3−2x2+6x−1=0 then the value of α(α2+α+1α2−α+1)+β(β2+β+1β2−β+1)+γ(γ2+γ+1γ2−γ+1)=

(α2+α+1α2−α+1)=α3−1(α−1)(α2−α+1) =α3−1α3−2α2+2α−1 1−6α+2α−1 =α3−1−4α=2α2−6α−4α =2α(α−3)−4α =(α−3)−2 =3−α2R.V= =∑3−α2=3(α+β+γ)−(α2+β2+γ2)2 =6−(4−2(6))2 =6+82 =7

If α,β,γ are roots of x3−2x2+6x−1=0 then the value of α(α2+α+1α2−α+1)+β(β2+β+1β2−β+1)+γ(γ2+γ+1γ2−γ+1)=

(α2+α+1α2−α+1)=α3−1(α−1)(α2−α+1) =α3−1α3−2α2+2α−1 1−6α+2α−1 =α3−1−4α=2α2−6α−4α =2α(α−3)−4α =(α−3)−2 =3−α2R.V= =∑3−α2=3(α+β+γ)−(α2+β2+γ2)2 =6−(4−2(6))2 =6+82 =7

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If $α, β, γ$ are roots of $x^{3} - 2 x^{2} + 6 x - 1 = 0$ then the value of $α (\frac{α^{2} + α + 1}{α^{2} - α + 1}) + β (\frac{β^{2} + β + 1}{β^{2} - β + 1}) + γ (\frac{γ^{2} + γ + 1}{γ^{2} - γ + 1}) =$

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Detailed Solution

$(\frac{α^{2} + α + 1}{α^{2} - α + 1}) = \frac{α^{3} - 1}{(α - 1) (α^{2} - α + 1)} = \frac{α^{3} - 1}{α^{3} - 2 α^{2} + 2 α - 1} 1 - 6 α + 2 α - 1 = \frac{α^{3} - 1}{- 4 α} = \frac{2 α^{2} - 6 α}{- 4 α} = \frac{2 α (α - 3)}{- 4 α} = \frac{(α - 3)}{- 2} = \frac{3 - α}{2}$