Rijul saini wrote: Suppose that the function $f:\mathbb{R}\to\mathbb{R}$ satisfies \[f(x^3 + y^3)=(x+y)(f(x)^2-f(x)f(y)+f(y)^2)\] for all $x,y\in\mathbb{R}$. Prove that $f(1996x)=1996f(x)$ for all $x\in\mathbb{R}$. This has surely already been posted. Let $P(x,y)$ be the assertion $f(x^3+y^3)=(x+y)(f(x)^2-f(x)f(y)+f(y)^2)$ Let $A=\{a\in\mathbb R$ such that $f(ax)=af(x)$ $\forall x\}$. Obviously $1\in A$ $P(0,0)$ $\implies$ $f(0)=0$ (and so $0\in A$) $P(x,0)$ $\implies$ $f(x^3)=xf(x)^2$ and $f(x)$ and $x$ always have same signs. Let $a\in A$ : $P(ax,x)$ $\implies$ $f((a^3+1)x^3)=(ax+x)(a^2f(x)^2-af(x)^2+f(x)^2)$ $=(a^3+1)xf(x)^2=(a^3+1)f(x^3)$ And so $a\in A$ $\implies$ $a^3+1\in A$ (*) Let $x\ne 0$ and $a\in A$ : $P(\sqrt[3]ax,0)$ $\implies$ $f(ax^3)=\sqrt[3]axf(\sqrt[3]x)^2$ and since $f(ax^3)=af(x^3)=axf(x)^2$ : $(\sqrt[3]af(x))^2=f(\sqrt[3]ax)^2$ and, since $f(x)$ and $x$ always have same signs : $f(\sqrt[3]ax)=\sqrt[3]af(x)$ and so $a\in A$ $\implies$ $\sqrt[3]a\in A$ (**) So, using (**) and then (*), we get $a\in A$ $\implies$ $a+1\in A$ and, since $1\in A$, we get $1996\in A$ Q.E.D.

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Let $P(x,y)$ denote $f(x^3+y^3)=(x+y)(f(x)^2-f(x)f(y)+f(y)^2).$ Clearly $f(0)=0$ and $P(\sqrt[3]{x},0)$ implies $f(x)=\sqrt[3]x f(\sqrt[3] x)^2$, in particular $f(x)\geqslant 0$ for $x\geqslant 0$ and vice-versa. We prove by induction that $f(nx)=nf(x)$ for all $n\in \mathbb N$. True for $n=1$, assume true for some $k$. Then $P(\sqrt[3]kx,0)$ implies $f(\sqrt[3]kx)=\sqrt[3]kf(x)$. And $P(\sqrt[3]{x},\sqrt[3]{kx})$ implies $f(x(k+1))=f(x)(k+1)$, as desired.