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Question Number 40910 by rahul 19 last updated on 29/Jul/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18
	$let the required eqn is (x−A)(x−B)(x−C)=0 A=α+β+γ−2γ A=((−b)/a)−2γ B=((−b)/a)−2α C=((−b)/a)−−2β (x+(b/a)+2γ)(x+(b/2)+2α)(x+(b/a)+2β)=0 (t+2γ)(t+2α)(t+2β)=0 t=x+(b/a) (t+2γ){t^2 +2t(α+β)+4αβ}=0 t^3 +2t^2 (α+β)+4tαβ+2t^2 γ+4t(αγ+βγ)+8αβγ=0 t^3 +2t^2 (α+β+γ)+4t(αβ+αγ+βγ)+8αβγ=0 t^3 +2t^2 (((−b)/a))+4t((c/a))+8(((−d)/a))=0 (x+(b/a))^3 +2(x+(b/a))^2 (((−b)/a))+4(x+(b/a))+8(((−d)/a))=0 rest portion you simplyfy....$
	$l e t t h e r e q u i r e d e q n i s$ $(x - A) (x - B) (x - C) = 0$ $A = α + β + γ - 2 γ$ $A = \frac{- b}{a} - 2 γ B = \frac{- b}{a} - 2 α C = \frac{- b}{a} - - 2 β$ $(x + \frac{b}{a} + 2 γ) (x + \frac{b}{2} + 2 α) (x + \frac{b}{a} + 2 β) = 0$ $(t + 2 γ) (t + 2 α) (t + 2 β) = 0$ $t = x + \frac{b}{a}$ $(t + 2 γ) {t^{2} + 2 t (α + β) + 4 α β} = 0$ $t^{3} + 2 t^{2} (α + β) + 4 t α β + 2 t^{2} γ + 4 t (α γ + β γ) + 8 α β γ = 0$ $t^{3} + 2 t^{2} (α + β + γ) + 4 t (α β + α γ + β γ) + 8 α β γ = 0$ $t^{3} + 2 t^{2} (\frac{- b}{a}) + 4 t (\frac{c}{a}) + 8 (\frac{- d}{a}) = 0$ ${(x + \frac{b}{a})}^{3} + 2 {(x + \frac{b}{a})}^{2} (\frac{- b}{a}) + 4 (x + \frac{b}{a}) + 8 (\frac{- d}{a}) = 0$ $r e s t p o r t i o n y o u s i m p l y f y . . . .$
	Commented by rahul 19 last updated on 30/Jul/18

	$thanks sir .$
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