x3−2x2+3x−4=0 So, α+β+γ=2αβ+βγ+γα=+3 αβγ=4 thus, α2+β2+γ2=(α+β+γ)2 −2(αβ+βγ+γα) =4−2(3)=−2 and α2β2+β2γ2+γ2α2 =(αβ)2+(βγ)2+(γα)2 ∵α2β2+β2γ2+γ2α2 =(αβ+βγ+γα)2−2αβγ(α+β+γ) =(3)2−2(4)(2)=9−16=−7 and α2β2γ2=(αβγ)2=42=16 Hence, the cubic equation with roots α2,β2,γ2 is x3−(α2+β2+γ2)x2+(α2β2+β2γ2+γ2α2)x−α2β2γ2 =x3−(−2)x2+(−7)x−16x−α2β2γ2 =x3+2x2−7x−16=0
