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Determine all $f : \mathbb R \rightarrow \mathbb R$ such that $$f(x^3 + y^3) = x^2\,f(x) + y\,f(y^2)$$ for all $x,y\in\mathbb{R}$.

I put $x = y = 0$ then $x = 0$ and $y = 0$ respectively. This implies Cauchy’s Functional Equation , but I can’t continue after this.

Batominovski
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Oleg Farenskiy
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1 Answers1

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By setting $x$ and $y$ to be $0$, we obtain $f(0)=0$. Now, with $y:=0$, we get $$f(x^3)=x^2\,f(x)\tag{*}$$ for all $x\in\mathbb{R}$. Likewise, with $x:=0$, we get $$f(y^3)=y\,f(y^2)\tag{#}$$ for all $y\in\mathbb{R}$. This shows that $$x\,f(x^2)=f(x^3)=x^2\,f(x)$$ for every $x\in\mathbb{R}$. Thus, along with $f(0)=0$, we conclude that $$f(x^2)=x\,f(x)\tag{$\star$}$$ for all $x\in\mathbb{R}$.

Now, we have by (*) and (#) that $$\begin{align}f(x+y)&=f\Big((\sqrt[3]{x})^3+(\sqrt[3]{y})^3\Big)\\&=(\sqrt[3]{x})^2\,f\big(\sqrt[3]{x}\big)+\sqrt[3]{y}\,f\big((\sqrt[3]{y})^2\big)\\&=f\big((\sqrt[3]{x})^3\big)+f\big((\sqrt[3]{y})^3\big)=f(x)+f(y)\tag{\$}\end{align}$$ for every $x,y\in\mathbb{R}$. From ($\star$) and (\$), we have $$f\big((x+1)^2\big)=(x+1)\,f(x+1)=(x+1)\,\big(f(x)+f(1)\big)$$ for any $x\in\mathbb{R}$. However, from ($\star$) and (\$), we also have $$\begin{align}f\big((x+1)^2\big)&=f(x^2+2x+1)\\&=f(x^2)+2\,f(x)+f(1)=x\,f(x)+2\,f(x)+f(1)\end{align}$$ for any $x\in\mathbb{R}$. That is, $$(x+1)\,\big(f(x)+f(1)\big)=x\,f(x)+2\,f(x)+f(1)$$ for any $x\in\mathbb{R}$, whence $$f(x)=f(1)\,x$$ for all $x\in\mathbb{R}$.

Batominovski
  • 49,629