A Functional Equation from Türkiye 🇹🇷

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[MortezaSabzian-dbsl]

f(x)+f(1/3_√(1-x³))=x³
Let and set x³→x
f(3_√x)+f(1/3√(1-x)=x
Set and let f(3_√u)=g(u)
g(x)+g(1/(1-x)=x
This functional equation has been solved many times
f(x)=(x⁹-x³+1)/2x³(x³-1)
f(-1)=1/4
f(∞)=∞

[MortezaSabzian-dbsl]

To a small generalization. If we have an equation like this;

where the function u^(-1)(x) is the inverse of u(x).
Let u^(-1)(x)→x
f(u(x))+f(u(1/(1-x)))=v(u(x))
and let f(u(x))=g(x)
With the following few substitutions we arrive at a system of equations
g(x)+g(1/(1-x))=v(u(x))
x→1/(1-x)

x→1/(1-x)
g((x-1)/x)+g(x)=v(u((x-1)/x))
As result we have;
1)g(x)+g(1/(1-x))=v(u(x))




We subtract equation 2 from 3 and add the result to equation 1;

So

For example
u(x)=³√x
v(x)=x³

[scottleung]

I went down the route of the second method - like you said, I didn't get anywhere far.

[giuseppepapari]

2:53. I am confused here: 1/ [ 1/ cube_root(1/2)] should be cube_root(1/2). Or do I miss something?

[paulkarch]

I tried this substitution for x. ((x^3 - 1)/x^3)^(1/3) and got f((1 - 1/x^3)^(1/3))) + f(x) = 1 - 1/x^3 but not sure if this leads anywhere.