If α,β,γ be the roots of x3 + x2 - 4x + 7 = 0, then find a cubic equation whose roots are β+γ,γ+α,α+β

Let y=β+γ=α+β+γ−α=−1−α α=−(y+1)Since α is a root of the given equation, soγ3+γ2+4α+7=0−(y+1)3+(y+1)2−4(y+1)+7=0(y+1)3−(y+1)2+4(y+1)−7=0Hence, the equation is(x+1)3−(x+1)2+4(x+1)−7=0

If α,β,γ be the roots of x3 + x2 - 4x + 7 = 0, then find a cubic equation whose roots are β+γ,γ+α,α+β

Let y=β+γ=α+β+γ−α=−1−α α=−(y+1)Since α is a root of the given equation, soγ3+γ2+4α+7=0−(y+1)3+(y+1)2−4(y+1)+7=0(y+1)3−(y+1)2+4(y+1)−7=0Hence, the equation is(x+1)3−(x+1)2+4(x+1)−7=0

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If $α, β, γ$ be the roots of x³ + x² - 4x + 7 = 0, then find a cubic equation whose roots are $β + γ, γ + α, α + β$

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x=-0

$x + 1$

$(x + 1)^{3} - (x + 1)^{2} + 4 (x + 1) - 7 = 0$

answer is D.

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Detailed Solution

Let

$y = β + γ = α + β + γ - α = - 1 - α α = - (y + 1)$

Since $α$ is a root of the given equation, so

$\begin{array}{l} γ^{3} + γ^{2} + 4 α + 7 = 0 \\ - (y + 1)^{3} + (y + 1)^{2} - 4 (y + 1) + 7 = 0 \\ (y + 1)^{3} - (y + 1)^{2} + 4 (y + 1) - 7 = 0 \end{array}$