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Question-211881




Question Number 211881 by Spillover last updated on 23/Sep/24
Answered by som(math1967) last updated on 23/Sep/24
 α+β+γ=1,αβ+βγ+γα=−2   αβγ=−1   α^2 +β^2 +γ^2 =1+4=5  ^3 (√(α^6 /(α^2 β^2 γ^2 )))+^3 (√(β^6 /(α^2 β^2 γ^2 )))+^3 (√(γ^6 /(α^2 β^2 γ^2 )))  =(α^2 /1)+(β^2 /1)+(γ^2 /1)  =(α^2 +β^2 +γ^2 )=5
$$\:\alpha+\beta+\gamma=\mathrm{1},\alpha\beta+\beta\gamma+\gamma\alpha=−\mathrm{2} \\ $$$$\:\alpha\beta\gamma=−\mathrm{1} \\ $$$$\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} =\mathrm{1}+\mathrm{4}=\mathrm{5} \\ $$$$\:^{\mathrm{3}} \sqrt{\frac{\alpha^{\mathrm{6}} }{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} }}+^{\mathrm{3}} \sqrt{\frac{\beta^{\mathrm{6}} }{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} }}+^{\mathrm{3}} \sqrt{\frac{\gamma^{\mathrm{6}} }{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} }} \\ $$$$=\frac{\alpha^{\mathrm{2}} }{\mathrm{1}}+\frac{\beta^{\mathrm{2}} }{\mathrm{1}}+\frac{\gamma^{\mathrm{2}} }{\mathrm{1}} \\ $$$$=\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)=\mathrm{5} \\ $$
Commented by Spillover last updated on 23/Sep/24
correct.thanks
$${correct}.{thanks} \\ $$

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