Given α+β+γ=2,α2+β2+γ2=6 and α3+β3+γ3=8, then the equation whose roots are α,β,γ is:
A
x3−2x2+x−2=0
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B
x3−2x2−x+2=0
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C
x3−2x2+2x−4=0
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D
x3−2x2+3x−6=0
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Solution
The correct option is Ax3−2x2−x+2=0 We know 2(αβ+βγ+γα)=(α+β+γ)2−(α2+β2+γ2)=4−6=−2⇒αβ+βγ+γα=−1 And α3+β3+γ3−3αβγ=(α+β+γ)(α2+β2+γ2−αβ−βγ−γα)⇒3αβγ=8−2(6+1)=−6⇒αβγ=−2 So, the required equation is x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ=0⇒x3−2x2−x+2=0