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Question Number 186105 by HeferH last updated on 01/Feb/23
a^2 + 1 = a a^(12) + a^6 + 1 = ?
a2+1=aa12+a6+1=?
Answered by Rasheed.Sindhi last updated on 01/Feb/23
a^2 +1=a; a^(12) +a^6 +1=? a^2 =a−1 a^3 =a^2 −a=(a−1)−a=−1 ▶a^(12) +a^6 +1=(a^3 )^4 +(a^3 )^2 +1 =(−1)^4 +(−1)^2 +1=3
a2+1=a;a12+a6+1=?a2=a−1a3=a2−a=(a−1)−a=−1▸a12+a6+1=(a3)4+(a3)2+1=(−1)4+(−1)2+1=3
Answered by Rasheed.Sindhi last updated on 01/Feb/23
a^2 −a+1=0⇒a+(1/a)=1 ; a≠0 (a+(1/a))^2 =1^2 ⇒a^2 +(1/a^2 )=−1 (a^2 +(1/a^2 ))^3 =(−1)^3 a^6 +(1/a^6 )+3(a^2 +(1/a^2 ))=−1 a^6 +(1/a^6 )+3(−1)=−1⇒a^6 +(1/a^6 )=2 a^(12) −2a^6 +1=0 (a^6 −1)^2 =0 a^6 =1 a^(12) +a^6 +1=a^6 (a^6 +(1/a^6 )+1) =(1)(2+1)=3
a2−a+1=0⇒a+1a=1;a≠0(a+1a)2=12⇒a2+1a2=−1(a2+1a2)3=(−1)3a6+1a6+3(a2+1a2)=−1a6+1a6+3(−1)=−1⇒a6+1a6=2a12−2a6+1=0(a6−1)2=0a6=1a12+a6+1=a6(a6+1a6+1)=(1)(2+1)=3
Answered by Rasheed.Sindhi last updated on 01/Feb/23
a^2 + 1 = a; a^(12) + a^6 + 1 = ? (a^2 +1)^2 =a^2 a^4 +2a^2 +1=a^2 a^4 +1=−a^2 (a^4 +1)^3 =(−a^2 )^3 a^(12) +1+3a^4 (a^4 +1)=−a^6 a^(12) +1+3a^4 (−a^2 )=−a^6 a^(12) −2a^6 +1=0 (a^6 −1)^2 =0 a^6 =1 ▶a^(12) +a^6 +1=(a^6 )^2 +a^6 +1 =(1)^2 +1+1=3
a2+1=a;a12+a6+1=?(a2+1)2=a2a4+2a2+1=a2a4+1=−a2(a4+1)3=(−a2)3a12+1+3a4(a4+1)=−a6a12+1+3a4(−a2)=−a6a12−2a6+1=0(a6−1)2=0a6=1▸a12+a6+1=(a6)2+a6+1=(1)2+1+1=3
Commented by HeferH last updated on 01/Feb/23
(thank you)^3 !
(thankyou)3!

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