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

α+β+γ=0 ∑1α+β−γ=∑1−2γ=1-2α+1-2β+1-2γ =−12(∑αβαβγ) = q2r

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

α+β+γ=0 ∑1α+β−γ=∑1−2γ=1-2α+1-2β+1-2γ =−12(∑αβαβγ) = q2r

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

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$\frac{q}{2 r}$

$\frac{q}{r}$

$\frac{2 q}{r}$

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

answer is A.

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$α + β + γ = 0$

$\sum \frac{1}{α + β - γ} = \sum \frac{1}{- 2 γ}$ = $\frac{1}{- 2 α} + \frac{1}{- 2 β} + \frac{1}{- 2 γ}$

$= \frac{- 1}{2} (\frac{\sum α β}{α β γ})$

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

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

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