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a-8-a-4-1-a-6-1-solves-

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Question Number 11477 by @ANTARES_VY last updated on 26/Mar/17
((a^8 +a^4 +1)/(a^6 +1)). solves....
a8+a4+1a6+1.solves….
Answered by sm3l2996 last updated on 26/Mar/17
a^8 +a^4 +1=a^2 (a^6 +1)+a^4 −a^2 +1 ((a^8 +a^4 +1)/(a^6 +1))=a^2 +((a^4 −a^2 +1)/(a^6 +1)) ((a^4 −a^2 +1)/(a^6 +1))=((a^4 −a^2 +1)/((a^2 +1)(a^4 −a^2 +1)))=(1/(a^2 +1)) ((a^8 +a^4 +1)/(a^6 +1))=a^2 +(1/(a^2 +1))
a8+a4+1=a2(a6+1)+a4−a2+1a8+a4+1a6+1=a2+a4−a2+1a6+1a4−a2+1a6+1=a4−a2+1(a2+1)(a4−a2+1)=1a2+1a8+a4+1a6+1=a2+1a2+1
Commented by @ANTARES_VY last updated on 27/Mar/17
Thenks.
Thenks.
Commented by @ANTARES_VY last updated on 27/Mar/17
Thenks.
Thenks.
Commented by @ANTARES_VY last updated on 27/Mar/17
Thenks.
Thenks.
Commented by @ANTARES_VY last updated on 27/Mar/17
Thenks.
Thenks.
Commented by sma3l2996 last updated on 27/Mar/17
you welcome
youwelcome
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Mar/17
a^2 =p⇒((p^4 +p^2 +1)/(p^3 +1))=(((p^6 −1)/(p^2 −1))/(p^3 +1))=(((p^3 +1)(p^3 −1))/((p^3 +1)(p^2 −1))) =(((p−1)(p^2 +p+1))/((p−1)(p+1)))=((p^2 +p+1)/(p+1))=((a^4 +a^2 +1)/(a^2 +1)) (((a^2 +1)^2 −a^2 )/(a^2 +1))=(((a^2 +a+1)(a^2 −a+1))/(a^2 +1))
a2=p⇒p4+p2+1p3+1=p6−1p2−1p3+1=(p3+1)(p3−1)(p3+1)(p2−1)=(p−1)(p2+p+1)(p−1)(p+1)=p2+p+1p+1=a4+a2+1a2+1(a2+1)2−a2a2+1=(a2+a+1)(a2−a+1)a2+1

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