take $y=\frac{1-x}{1+x}$, then $f^2(\frac{1-x}{1+x})f(x)=64\frac{1-x}{1+x}$. It give $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}.$

Rust wrote: take $y=\frac{1-x}{1+x}$, then $f^2(\frac{1-x}{1+x})f(x)=64\frac{1-x}{1+x}$. It give $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}.$ With $x_0$ so that $f(x_0) = 0$ or $f^2(\frac{1-x_0}{1+x_0}) = 0$ then you can't solve this system by dividing them to find $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}$. And how do you dispose continue?

dyta wrote: Rust wrote: take $y=\frac{1-x}{1+x}$, then $f^2(\frac{1-x}{1+x})f(x)=64\frac{1-x}{1+x}$. It give $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}.$ With $x_0$ so that $f(x_0) = 0$ or $f^2(\frac{1-x_0}{1+x_0}) = 0$ then you can't solve this system by dividing them to find $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}$. And how do you dispose continue? Rust is right, as often, although rather short, as always Let $A=\mathbb R\setminus \{-1,0,1\}$ Let $u(x)=\frac {1-x}{1+x}$ function from $A\to A$. Notice that $u(u(x))=x$ $\forall x\in A$ Let $P(x)$ be the assertion $f(x)^2f(u(x))=64x$ $\forall x\in A$ Squaring $P(x)$, we get $f(x)^4f(u(x))^2=2^{12}x^2$ $\forall x\in A$ $x\in A\implies u(x)\in A$ and so $P(u(x))$ $\implies$ $f(u(x))^2f(x)=64u(x)$ $x\in A$ $\implies$ $u(x)\ne 0$ and so $f(u(x))^2f(x)\ne 0$ and so we can divide the first line by the second and we get : $f(x)^3=2^6\frac{x^2}{u(x)}$ $\forall x\in A$ And so $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}$ $\forall x\in A$, as Rust previously said. And it just remains to define $f(-1),f(0)$ and $f(1)$ as any values we want. And it's easy to check back that this mandatory form indeed is a solution. And, btw, I dont understand your phrase "how do you dispose continue" ?

Ok, I understood, we'll do when $u(x)=0$ now.

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pco wrote: dyta wrote: Rust wrote: take $y=\frac{1-x}{1+x}$, then $f^2(\frac{1-x}{1+x})f(x)=64\frac{1-x}{1+x}$. It give $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}.$ With $x_0$ so that $f(x_0) = 0$ or $f^2(\frac{1-x_0}{1+x_0}) = 0$ then you can't solve this system by dividing them to find $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}$. And how do you dispose continue? Rust is right, as often, although rather short, as always Let $A=\mathbb R\setminus \{-1,0,1\}$ Let $u(x)=\frac {1-x}{1+x}$ function from $A\to A$. Notice that $u(u(x))=x$ $\forall x\in A$ Let $P(x)$ be the assertion $f(x)^2f(u(x))=64x$ $\forall x\in A$ Squaring $P(x)$, we get $f(x)^4f(u(x))^2=2^{12}x^2$ $\forall x\in A$ $x\in A\implies u(x)\in A$ and so $P(u(x))$ $\implies$ $f(u(x))^2f(x)=64u(x)$ $x\in A$ $\implies$ $u(x)\ne 0$ and so $f(u(x))^2f(x)\ne 0$ and so we can divide the first line by the second and we get : $f(x)^3=2^6\frac{x^2}{u(x)}$ $\forall x\in A$ And so $f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}$ $\forall x\in A$, as Rust previously said. And it just remains to define $f(-1),f(0)$ and $f(1)$ as any values we want. And it's easy to check back that this mandatory form indeed is a solution. And, btw, I dont understand your phrase "how do you dispose continue" ? I think this observation makes this one liner. Note that 0 cannot be the range of f. Let's prove by contradiction. Suppose f(a)=0 then( f(a)^2)(f(1-a)/(1+a))=0 . But 64a cannot be 0 since 0 cannot be the domain for the equation given in the question

Note that if we replace $x=\frac{1-x}{1+x}$, we get: \[f\left(\frac{1-x}{1+x}\right)^2 f(x)=\frac{64(1-x)}{1+x}\]and this holds for $x\neq 0$, $x\neq 1$, $x\neq -1$. Now note that: \begin{align*} f\left(\frac{1-x}{1+x}\right)f(x)^2\left[f(x)-f\left(\frac{1-x}{1+x}\right)\right] &= f(x)\left[f\left(\frac{1-x}{1+x}\right)f(x)^2-f\left(\frac{1-x}{1+x}\right)^2 f(x)\right]\\ &= f(x)\left[64x-\frac{64(1-x)}{1+x}\right]\\ &= \frac{64f(x)(x^2+2x-1)}{x+1}. \end{align*}By replacing $f\left(\frac{1-x}{1+x}\right)f(x)^2$ in the left side of the equation we get: \[f(x)-f\left(\frac{1-x}{1+x}\right)=\frac{f(x)(x^2+2x-1)}{x(x+1)}\]\[f\left(\frac{1-x}{1+x}\right)=\frac{f(x)(1-x)}{x(1+x)}\]And finally, by replacing this in the original equation, we find that: \[\frac{f(x)^3(1-x)}{x(1+x)}=64x \implies f(x)=4\sqrt[3]{\frac{x^2(1+x)}{1-x}}\]