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Question
If α,β,γ are the roots of x3+2x2+3x+8=0, then (α+β)(β+γ)(γ+α)=
If α,β,γ are the roots of x3+2x2+3x+8=0, then (α+β)(β+γ)(γ+α)=
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Solution
The correct option is A 2
As α,β,γ are roots of x3+2x3+3x+8=0
∴α+β+γ=−2,αβ+βγ+γα=3,αβγ=−8
Now
(α+β)(β+γ)(γ+α)=(−2−γ)(−2−α)(−2−β)=−(2+γ)(2+α)(2+β)=−(2+γ)(4+2(α+β)+αβ)=−(8+4(α+β)+2αβ+4γ+2(α+β)γ+αβγ)=−(8+4(α+β+γ)+2(αβ+γβ+βα)+αβγ)=−(8+4(−2)+2(3)−8)=−(8−8+6−8)=2
As α,β,γ are roots of x3+2x3+3x+8=0
∴α+β+γ=−2,αβ+βγ+γα=3,αβγ=−8
Now
(α+β)(β+γ)(γ+α)=(−2−γ)(−2−α)(−2−β)=−(2+γ)(2+α)(2+β)=−(2+γ)(4+2(α+β)+αβ)=−(8+4(α+β)+2αβ+4γ+2(α+β)γ+αβγ)=−(8+4(α+β+γ)+2(αβ+γβ+βα)+αβγ)=−(8+4(−2)+2(3)−8)=−(8−8+6−8)=2
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