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f-x-1-x-x-6-1-27-f-x-

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Question Number 78794 by naka3546 last updated on 20/Jan/20
f(x + (1/x)) = ((x^6 + 1)/(27)) f(x) = ...
f(x+1x)=x6+127f(x)=…
Answered by mr W last updated on 20/Jan/20
t=x+(1/x) x^2 −tx+1=0 ⇒x=(1/2)(t±(√(t^2 −4))) x^2 =tx−1 x^3 =tx^2 −x=t(tx−1)−x=(t^2 −1)x−t x^6 =(t^2 −1)^2 x^2 −2t(t^2 −1)x+t^2 x^6 =(t^2 −1)^2 (tx−1)−2t(t^2 −1)x+t^2 x^6 =t(t^2 −3)[(t^2 −1)x−t]−1 x^6 +1=t(t^2 −3)[(t^2 −1)x−t] ((x^6 +1)/(27))=((t(t^2 −3))/(27))[(((t^2 −1)(t±(√(t^2 −4))))/2)−t] f(t)=((t(t^2 −3))/(27))[(((t^2 −1)(t±(√(t^2 −4))))/2)−t] ⇒f(x)=((x(x^2 −3))/(27))[(((x^2 −1)(x±(√(x^2 −4))))/2)−x]
t=x+1xx2−tx+1=0⇒x=12(t±t2−4)x2=tx−1x3=tx2−x=t(tx−1)−x=(t2−1)x−tx6=(t2−1)2x2−2t(t2−1)x+t2x6=(t2−1)2(tx−1)−2t(t2−1)x+t2x6=t(t2−3)[(t2−1)x−t]−1x6+1=t(t2−3)[(t2−1)x−t]x6+127=t(t2−3)27[(t2−1)(t±t2−4)2−t]f(t)=t(t2−3)27[(t2−1)(t±t2−4)2−t]⇒f(x)=x(x2−3)27[(x2−1)(x±x2−4)2−x]

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