If α,β,γ are the roots of x3+ax2+bx+c=0, then ∏(α+β−2γ)=
A
2a3−9ab+27c
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B
2a3+9ab+27c
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C
2a3−9ao−27c
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D
2a3+9ao−27c
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Solution
The correct option is A2a3−9ab+27c As α,β,γ are roots of x3+ax2+bx+c=0 Gives s1=α+β+γ=−a⇒α+β=−a−γs2=αβ+βγ+γα=bs3=αβγ=c Therefore, Π(α+β−2γ)=Π(−a−2γ)=−(a+3α)(a+3β)(a+3γ)=−(a3+3a2(α+β+γ)+3a(αβ+βγ+γα)+2+αβγ)=−(a3−3a3+9ab−27c)=2a3−9ab+27c