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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 ✓

x6−x4+1=0(x2)3−(x2)2+1=0x12+x22+x32=1x12x22+x22x32+x32x12=0x12x22x32=−1(x12+x22+x32)2=12x14+x24+x34+2(x12x22+x22x32+x32x12)=1x14+x24+x34+2×0=1⇒x14+x24+x34=1αk12=(αk6)2=(αk4−1)2=αk8−2αk4+1=αk2(αk4−1)−2αk4+1=αk6−αk2−2αk4+1=αk4−1−αk2−2αk4+1=−(αk2+αk4)∑6k=1αk12=−2∑3k=1(αk2+αk4)=−2(1+1)=−4✓

Commented by Rupesh123 last updated on 08/Nov/23

Perfect, sir!

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