If $α, β, γ$ are the roots of x³+x²+2x+ 3 = 0 then the equation whose roots $β + γ, γ + α, α + β$ , is

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$x^{3} + 2 x^{2} + 3 x - 1 = 0$

$x^{3} + 2 x^{2} + 3 x + 1 = 0$

$x^{3} + 2 x^{2} - 3 x - 1 = 0$

$x^{3} - 2 x^{2} + 3 x - 1 = 0$

answer is A.

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Detailed Solution

$α + β + γ = - 1 α β + β γ + γ α = 2 α β γ = - 3 (i) β + γ + γ + α + α + β = 2 (- 1) \Rightarrow - 2 (i i) = (β + γ) (γ + α) + (γ + α) (α + β) + (α + β) (β + γ) = (β γ + β α + γ^{2} + γ α) + (α γ + γ β + α^{2} + α β) + (α β + α γ + β^{2} + β γ) = 3 (α β + β γ + γ α) + α^{2} + β^{2} + γ^{2} = 3 (α β + β γ + γ α) + (α + β + γ)^{2} - 2 (α β + β γ + γ α) = 3 (i i i) = (β + γ) (γ + α) (α + β) = α β γ + α^{2} β + α γ^{2} + α^{2} γ + β^{2} γ + α β^{2} + β γ^{2} + α β γ = α (β γ + γ α + α β) + β (β γ + γ α + α β) + γ (β γ + γ α + α β) - α β γ = (α + β + γ) (β γ + γ α + α β) - α β γ = 1 F r o m (i), (i i) a n d (i i i) w e g e t, x^{3} - x^{2} (β + γ + γ + α + α + β) + x ((β + γ) (γ + α) + (γ + α) (α + β) + (α + β) (β + γ)) - ((β + γ) (γ + α) (α + β)) = 0 \Rightarrow x^{3} - (- 2) x^{2} + (3) x - (1) = 0 \Rightarrow x^{3} + 2 x^{2} + 3 x - 1 = 0$