If α,β,γ are the roots of 2x3 – 2x – 1 = 0 the ∑αβ2=

If α,β,γ are roots of 2x3−2x−1=0α+β+γ=0, αβ+βγ+γα=−1, αβγ= 12∑(αβ)2=(αβ)2+(βγ)2+(γα)2=(αβ+βγ+αγ)2−2(αβ2γ+αβγ2+α2βγ)=(αβ+βγ+αγ)2−2αβγ(α+β+γ)=1−2(12)0=1

If α,β,γ are the roots of 2x3 – 2x – 1 = 0 the ∑αβ2=

If α,β,γ are roots of 2x3−2x−1=0α+β+γ=0, αβ+βγ+γα=−1, αβγ= 12∑(αβ)2=(αβ)2+(βγ)2+(γα)2=(αβ+βγ+αγ)2−2(αβ2γ+αβγ2+α2βγ)=(αβ+βγ+αγ)2−2αβγ(α+β+γ)=1−2(12)0=1

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If $α, β, γ$ are the roots of $2 x^{3} - 2 x - 1 = 0$ the $\sum {(αβ)}^{2} =$

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Detailed Solution

$\begin{array}{l} I f α, β, γ a r e r o o t s o f 2 x^{3} - 2 x - 1 = 0 \\ α + β + γ = 0, α β + β γ + γ α = - 1, α β γ = \frac{1}{2} \\ \sum {(α β)}^{2} = {(α β)}^{2} + {(β γ)}^{2} + {(γ α)}^{2} \\ = {(α β + β γ + α γ)}^{2} - 2 (α β^{2} γ + α β γ^{2} + α^{2} β γ) \\ = {(α β + β γ + α γ)}^{2} - 2 α β γ (α + β + γ) \\ = 1 - 2 (\frac{1}{2}) 0 \\ = 1 \end{array}$