If α,β,γ are the roots of x3 + 2x2 – 3x – 1 = 0 then α−2+β−2+γ−2=

α+β+γ=−2,αβ++βγ+γα=−3,αβγ=11α2+1β2+1γ2=αβ2+β2γ2+γ2α2α2β2γ2=(αβ+βγ+γα)2−2αβγ(α+β+γ)(αβγ)2=9−2(−2)1=13

If α,β,γ are the roots of x3 + 2x2 – 3x – 1 = 0 then α−2+β−2+γ−2=

α+β+γ=−2,αβ++βγ+γα=−3,αβγ=11α2+1β2+1γ2=αβ2+β2γ2+γ2α2α2β2γ2=(αβ+βγ+γα)2−2αβγ(α+β+γ)(αβγ)2=9−2(−2)1=13

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If $α, β, γ$ are the roots of x³ + 2x² – 3x – 1 = 0 then $α^{- 2} + β^{- 2} + γ^{- 2} =$

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Detailed Solution

$\begin{array}{l} α + β + γ = - 2, αβ + + βγ + γα = - 3, αβγ = 1 \\ \frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{γ^{2}} = \frac{{αβ}^{2} + β^{2} γ^{2} + γ^{2} α^{2}}{α^{2} β^{2} γ^{2}} \\ = \frac{(αβ + βγ + γα)^{2} - 2 αβγ (α + β + γ)}{(αβγ)^{2}} \\ = \frac{9 - 2 (- 2)}{1} = 13 \end{array}$