If α,β,γ are the roots of x3 + qx + r = 0 then 1α+β-γ+1β+γ-α+1γ+α-β=

If α,β,γ are roots of x3+qx+r=0α+β+γ =0, αβ+βγ+γα=q, αβγ=−r∴ 1α+β−γ+1β+γ−α+1γ+α−β=1−γ−γ+1−α−α+1−β−β= −12(1α+1β+1γ)=−12(αβ+βγ+γααβγ)=−12(q−r)=q2r

If α,β,γ are the roots of x3 + qx + r = 0 then 1α+β-γ+1β+γ-α+1γ+α-β=

If α,β,γ are roots of x3+qx+r=0α+β+γ =0, αβ+βγ+γα=q, αβγ=−r∴ 1α+β−γ+1β+γ−α+1γ+α−β=1−γ−γ+1−α−α+1−β−β= −12(1α+1β+1γ)=−12(αβ+βγ+γααβγ)=−12(q−r)=q2r

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

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$\frac{2 q}{r}$

$\frac{q}{2 r}$

$- \frac{2 q}{r}$

$\frac{q}{r}$

answer is A.

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Detailed Solution

$\begin{array}{l} I f α, β, γ a r e r o o t s o f x^{3} + q x + r = 0 \\ α + β + γ = 0, α β + β γ + γ α = q, α β γ = - r \\ ∴ \frac{1}{α + β - γ} + \frac{1}{β + γ - α} + \frac{1}{γ + α - β} = \frac{1}{- γ - γ} + \frac{1}{- α - α} + \frac{1}{- β - β} \\ = \frac{- 1}{2} (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}) \\ = \frac{- 1}{2} (\frac{α β + β γ + γ α}{α β γ}) \\ = \frac{- 1}{2} (\frac{q}{- r}) \\ = \frac{q}{2 r} \end{array}$