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

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

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

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

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

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Solution

The correct option is A $x^{3} + 2 x^{2} + 3 x - 1 = 0$
Given: $α, β, γ$ are the roots of $x^{3} + x^{2} + 2 x + 3 = 0$
To find the equation whose roots are $β + γ, γ + α, α + β$
Sol: We know,
$(i) α + β + γ = - 1 (i i) α β + β γ + γ α = 2 (i i i) α β γ = - 3$
Let $β + γ = a, γ + α = b, α + β = c$
Then $a, b, c$ are the roots of the desired polynomial. Now,
$(i^{'}) a + b + c = β + γ + γ + α + α + β = 2 (α + β + γ) = 2 (- 1) = - 2 (i i^{'}) a b + b c + c a = (β + γ) (γ + α) + (γ + α) (α + β) + (α + β) (β + γ) ⟹ = (β γ + β α + γ^{2} + γ α) + (α γ + γ β + α^{2} + α β) + (α β + α γ + β^{2} + β γ) ⟹ = 3 (α β + β γ + γ α) + α^{2} + β^{2} + γ^{2} ⟹ = 3 (α β + β γ + γ α) + (α + β + γ)^{2} - 2 (α β + β γ + γ α) ⟹ = (α β + β γ + γ α) + (α + β + γ)^{2} ⟹ = 2 + (- 1)^{2} = 3 (i i i^{'}) a b c = (β + γ) (γ + α) (α + β) ⟹ = α β γ + α^{2} β + α γ^{2} + α^{2} γ + β^{2} γ + α β^{2} + β γ^{2} + α β γ ⟹ = α (β γ + γ α + α β) + β (β γ + γ α + α β) + α γ^{2} + β γ^{2} + α β γ - α β γ ⟹ = α (β γ + γ α + α β) + β (β γ + γ α + α β) + γ (β γ + γ α + α β) - α β γ ⟹ = (α + β + γ) (β γ + γ α + α β) - α β γ ⟹ = (- 1) (2) - (- 3) = 1$
From (i'), (ii') and (iii') we get
$x^{3} - (a + b + c) x^{2} + (a b + b c + c a) x - a b c = 0 ⟹ x^{3} - (- 2) x^{2} + (3) x - (1) = 0 ⟹ x^{3} + 2 x^{2} + 3 x - 1 = 0$
is the required polynomial.

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