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Question-26942




Question Number 26942 by Mr eaay last updated on 31/Dec/17
Answered by mrW1 last updated on 31/Dec/17
α^3 −2α+2=0  β^3 −2β+2=0  γ^3 −2γ+2=0  α^3 +β^3 +γ^3 =2(α+β+γ)−6=2×0−6=−6  ⇒Answer (B)
$$\alpha^{\mathrm{3}} −\mathrm{2}\alpha+\mathrm{2}=\mathrm{0} \\ $$$$\beta^{\mathrm{3}} −\mathrm{2}\beta+\mathrm{2}=\mathrm{0} \\ $$$$\gamma^{\mathrm{3}} −\mathrm{2}\gamma+\mathrm{2}=\mathrm{0} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} =\mathrm{2}\left(\alpha+\beta+\gamma\right)−\mathrm{6}=\mathrm{2}×\mathrm{0}−\mathrm{6}=−\mathrm{6} \\ $$$$\Rightarrow{Answer}\:\left({B}\right) \\ $$

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