Question Number 11477 by @ANTARES_VY last updated on 26/Mar/17

((a^8 +a^4 +1)/(a^6 +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))

Commented by@ANTARES_VY last updated on 27/Mar/17

Thenks.

Commented by@ANTARES_VY last updated on 27/Mar/17

Thenks.

Commented by@ANTARES_VY last updated on 27/Mar/17

Thenks.

Commented by@ANTARES_VY last updated on 27/Mar/17

Thenks.

Commented bysma3l2996 last updated on 27/Mar/17

you welcome

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))