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Question Number 78794 by naka3546 last updated on 20/Jan/20

$$f(x+\frac{1}{x})=\frac{x^6+1}{27}$$$$f\left(x\right)=...$$

Answered by mr W last updated on 20/Jan/20

$$t=x+\frac{1}{x}$$$$x^2-tx+1=0$$$$\Rightarrow x=\frac{1}{2}(t\pm \sqrt{t^2-4})$$$$x^2=tx-1$$$$x^3={tx}^2-x=t(tx-1)-x=(t^2-1)x-t$$$$x^6={(t^2-1)}^2{x}^2-2t(t^2-1)x+t^2$$$$x^6={(t^2-1)}^2(tx-1)-2t(t^2-1)x+t^2$$$$x^6=t(t^2-3)[(t^2-1)x-t]-1$$$$x^6+1=t(t^2-3)[(t^2-1)x-t]$$$$\frac{x^6+1}{27}=\frac{t(t^2-3)}{27}[\frac{(t^2-1)(t\pm \sqrt{t^2-4})}{2}-t]$$$$f\left(t\right)=\frac{t(t^2-3)}{27}[\frac{(t^2-1)(t\pm \sqrt{t^2-4})}{2}-t]$$$$\Rightarrow f\left(x\right)=\frac{x(x^2-3)}{27}[\frac{(x^2-1)(x\pm \sqrt{x^2-4})}{2}-x]$$

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