1987 OIM Problems/Problem 1

Problem

Find all $f(x)$ such that: \[\left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x\] for $x \ne 0$, $x \ne 1$, $x \ne -1$,

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

We have the following equations: \[(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x\] \[(2) \left[ f\left(\frac{1-x}{1+x}\right) \right]^2f\left( x \right)=64 \cdot \frac{1-x}{1+x}\] Multiplying $(1)$ and $(2)$, we have \[(3) f(x)f\left(\frac{1-x}{1+x}\right) = 16 \cdot \sqrt[3]{x\cdot \frac{1-x}{1+x}}\] Dividing $(1)$ by $(3)$ gives \[f(x) = 4 \cdot \sqrt[3]{x^2\cdot \frac{1+x}{1-x}}\] Checking to see if it works… \[(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right) = 64 \cdot \sqrt[3]{x^4 \cdot \left(\frac{1+x}{1-x}\right)^2 \cdot \left(\frac{1-x}{1+x}\right)^2 \cdot \frac{1}{x}} = 64x\] ~Archieguan

See also

https://www.oma.org.ar/enunciados/ibe2.htm