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Question Number 131573 by mathlove last updated on 06/Feb/21

$$iff(x+\frac{1}{x})=x^3+\frac{1}{x^3}thenfaind$$$$f\left(\frac{1}{x}\right)=?$$

Answered by rs4089 last updated on 06/Feb/21

$$f(x+\frac{1}{x})=x^3+\frac{1}{x^3}$$$$f(x+\frac{1}{x})=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)$$$$f(x+\frac{1}{x})=(x+\frac{1}{x})\{{(x+\frac{1}{x})}^2-2-1\}$$$$f(x+\frac{1}{x})=(x+\frac{1}{x})\{{(x+\frac{1}{x})}^2-3\}$$$$nowputx+\frac{1}{x}=t$$$$f\left(t\right)=t(t^2-3)$$$$orf\left(x\right)=x(x^2-3)$$$$sof\left(\frac{1}{x}\right)=\frac{1}{x}(\frac{1}{x^2}-3)=\frac{1-3x^2}{x^3}$$$$$$

Commented by mathlove last updated on 06/Feb/21

$$tanks$$

Answered by mathmax by abdo last updated on 06/Feb/21

$$\mathrm{we}\mathrm{have}{\mathrm{a}}^3+{\mathrm{b}}^3={(\mathrm{a}+\mathrm{b})}^3-3\mathrm{ab}(\mathrm{a}+\mathrm{b})\Rightarrow$$$$\mathrm{f}(\mathrm{x}+\frac{1}{\mathrm{x}})={(\mathrm{x}+\frac{1}{\mathrm{x}})}^3-3(\mathrm{x}+\frac{1}{\mathrm{x}})\mathrm{let}\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow \mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^3-3\mathrm{t}\Rightarrow$$$$\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{1}{{\mathrm{x}}^3}-\frac{3}{\mathrm{x}}=\frac{{\mathrm{x}}^2-3}{{\mathrm{x}}^3}$$

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