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Question Number 169842 by mathlove last updated on 10/May/22

Answered by Rasheed.Sindhi last updated on 10/May/22

$$f^2\left(x\right).f\left(\frac{1-x}{1+x}\right)=x^2.............\left(\mathrm{A}\right)$$$${\left(\mathrm{A}\right)}^2:f^4\left(x\right).f^2\left(\frac{1-x}{1+x}\right)=x^4$$$$f^2\left(\frac{1-x}{1+x}\right)=\frac{x^4}{f{}^4\left(x\right)}$$$$\bullet x\to \frac{1-x}{1+x}$$$$\frac{1-x}{1+x}\to \frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}=\frac{1+x-1+x}{1+x+1-x}=\frac{2x}{2}=x$$$$\left(\mathrm{A}\right)\Rightarrow$$$$f^2\left(\frac{1-x}{1+x}\right).f\left(x\right)={\left(\frac{1-x}{1+x}\right)}^2$$$$\frac{x^4}{f{}^4\left(x\right)}.f\left(x\right)={\left(\frac{1-x}{1+x}\right)}^2$$$$\frac{x^4}{f{}^3\left(x\right)}={\left(\frac{1-x}{1+x}\right)}^2$$$$f{}^3\left(x\right)=x^4.{\left(\frac{1+x}{1-x}\right)}^2$$

Commented by mathlove last updated on 11/May/22

$$thankssir$$

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