Given that, α,β,γ, are the roots of the equation x3+2x2−3x−1=0 ⇒‌‌α+β+γ=−2 . . . (i) αβ+βγ+γα=−3 . . . (ii) and ‌‌αβγ=1 . . . (iii) On squaring equation (ii), we get ‌α2β2+β2γ2+γ2α2+2αβγ(α+β+γ)=9 ‌⇒α2β2+β2γ2+γ2α2=9−2‌ (1) ‌(−2)=13 Now ‌α−2+β−2+γ−2=‌