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Question-199733

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Question Number 199733 by Rupesh123 last updated on 08/Nov/23
Answered by mr W last updated on 08/Nov/23
x^6 −x^4 +1=0 (x^2 )^3 −(x^2 )^2 +1=0 x_1 ^2 +x_2 ^2 +x_3 ^2 =1 x_1 ^2 x_2 ^2 +x_2 ^2 x_3 ^2 +x_3 ^2 x_1 ^2 =0 x_1 ^2 x_2 ^2 x_3 ^2 =−1 (x_1 ^2 +x_2 ^2 +x_3 ^2 )^2 =1^2 x_1 ^4 +x_2 ^4 +x_3 ^4 +2(x_1 ^2 x_2 ^2 +x_2 ^2 x_3 ^2 +x_3 ^2 x_1 ^2 )=1 x_1 ^4 +x_2 ^4 +x_3 ^4 +2×0=1 ⇒x_1 ^4 +x_2 ^4 +x_3 ^4 =1 α_k ^(12) =(α_k ^6 )^2 =(α_k ^4 −1)^2 =α_k ^8 −2α_k ^4 +1 =α_k ^2 (α_k ^4 −1)−2α_k ^4 +1 =α_k ^6 −α_k ^2 −2α_k ^4 +1 =α_k ^4 −1−α_k ^2 −2α_k ^4 +1 =−(α_k ^2 +α_k ^4 ) Σ_(k=1) ^6 α_k ^(12) =−2Σ_(k=1) ^3 (α_k ^2 +α_k ^4 ) =−2(1+1)=−4 ✓
$${x}^{\mathrm{6}} −{x}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} −\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{1} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} {x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} {x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} {x}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} {x}_{\mathrm{2}} ^{\mathrm{2}} {x}_{\mathrm{3}} ^{\mathrm{2}} =−\mathrm{1} \\ $$$$\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{4}} +{x}_{\mathrm{2}} ^{\mathrm{4}} +{x}_{\mathrm{3}} ^{\mathrm{4}} +\mathrm{2}\left({x}_{\mathrm{1}} ^{\mathrm{2}} {x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} {x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} {x}_{\mathrm{1}} ^{\mathrm{2}} \right)=\mathrm{1} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{4}} +{x}_{\mathrm{2}} ^{\mathrm{4}} +{x}_{\mathrm{3}} ^{\mathrm{4}} +\mathrm{2}×\mathrm{0}=\mathrm{1} \\ $$$$\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{4}} +{x}_{\mathrm{2}} ^{\mathrm{4}} +{x}_{\mathrm{3}} ^{\mathrm{4}} =\mathrm{1} \\ $$$$ \\ $$$$\alpha_{{k}} ^{\mathrm{12}} =\left(\alpha_{{k}} ^{\mathrm{6}} \right)^{\mathrm{2}} =\left(\alpha_{{k}} ^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} =\alpha_{{k}} ^{\mathrm{8}} −\mathrm{2}\alpha_{{k}} ^{\mathrm{4}} +\mathrm{1} \\ $$$$\:\:\:=\alpha_{{k}} ^{\mathrm{2}} \left(\alpha_{{k}} ^{\mathrm{4}} −\mathrm{1}\right)−\mathrm{2}\alpha_{{k}} ^{\mathrm{4}} +\mathrm{1} \\ $$$$\:\:\:=\alpha_{{k}} ^{\mathrm{6}} −\alpha_{{k}} ^{\mathrm{2}} −\mathrm{2}\alpha_{{k}} ^{\mathrm{4}} +\mathrm{1} \\ $$$$\:\:\:=\alpha_{{k}} ^{\mathrm{4}} −\mathrm{1}−\alpha_{{k}} ^{\mathrm{2}} −\mathrm{2}\alpha_{{k}} ^{\mathrm{4}} +\mathrm{1} \\ $$$$\:\:\:=−\left(\alpha_{{k}} ^{\mathrm{2}} +\alpha_{{k}} ^{\mathrm{4}} \right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\alpha_{{k}} ^{\mathrm{12}} \:=−\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\left(\alpha_{{k}} ^{\mathrm{2}} +\alpha_{{k}} ^{\mathrm{4}} \right) \\ $$$$\:\:\:\:\:\:=−\mathrm{2}\left(\mathrm{1}+\mathrm{1}\right)=−\mathrm{4}\:\checkmark \\ $$
Commented by Rupesh123 last updated on 08/Nov/23
Perfect, sir!

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