To start,plugging $ y =0 $ we get $ f(x^3) = x^2f(x) $. Also, by putting $ y = -x $ we get $ f(x) = -f(-x) $, i.e, $ f $ is an odd function. Now if we take $ x =y $ we get $ f(2x^3 + x^2) = 2x^2f(x) + f(x^2) = 2f(x^3) + f(x^2) $.

No, by plugging $ y =0 $ you get $ f(x^3) = x^2f(x) +f(0)$; nothing yet points to $f(0)=0$.

mavropnevma wrote: No, by plugging $ y =0 $ you get $ f(x^3) = x^2f(x) +f(0)$; nothing yet points to $f(0)=0$. Just put $x=1$, and you can get $f(0)=0$

very nice functional equation!

Let $P(x,y)$ be the assertion that $f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy)$. $P(x,-x)$ yields $f(x)=-f(-x)$. From $P(1,0)$ we get that $f(0)=0$, hence from $P(x,0)$ we get $f(x^3)=x^2f(x)$. From $P(x,y)-P(x,-y)$ we get $f(y^3)+f(xy)=\dfrac{1}{2}\left ( f(x^3+y^3+xy)+f(y^3+xy-x^3) \right )$. Substituting back into $P(x,y)$ we get that $f(x^3+y^3+xy)=2f(x^3)+f(y^3+xy-x^3)\ (*)$ Let $a,b$ be two real numbers. Let $x=\sqrt[3]{a}$. The polynomial $P_b(\alpha)=\alpha^3+x\alpha-b$ has odd degree, hence at least one real root. Let $y$ be one of the real roots. Putting $x$ and $y$ in $(*)$, we get that $f(a+b)=2f(a)+f(b-a),\ \forall a,b\in \mathbb{R}$. Taking $a=b$, we get $f(2a)=2f(a)$, so \[f(x+y)=f(x)+f(y),\ \forall x,y\in \mathbb{R}\] So $f(x+y)=f(x)+f(y)$ and $f(x^3)=x^2f(x) (**)$. Taking $x+1$ and $x-1$ in $(**)$ and summing up the two relations, we get that $2x^2f(x)+6f(x)=f(x)(2x^2+2)+4xf(1)$, i.e. $f(x)=xf(1)$.

Aiscrim wrote: Let $P(x,y)$ be the assertion that $f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy)$. $P(x,-x)$ yields $f(x)=-f(-x)$. From $P(1,0)$ we get that $f(0)=0$, hence from $P(x,0)$ we get $f(x^3)=x^2f(x)$. From $P(x,y)-P(x,-y)$ we get $f(y^3)+f(xy)=\dfrac{1}{2}\left ( f(x^3+y^3+xy)+f(y^3+xy-x^3) \right )$. Substituting back into $P(x,y)$ we get that $f(x^3+y^3+xy)=2f(x^3)+f(y^3+xy-x^3)\ (*)$ Let $a,b$ be two real numbers. Let $x=\sqrt[3]{a}$. The polynomial $P_b(\alpha)=\alpha^3+x\alpha-b$ has odd degree, hence at least one real root. Let $y$ be one of the real roots. Putting $x$ and $y$ in $(*)$, we get that $f(a+b)=2f(a)+f(b-a),\ \forall a,b\in \mathbb{R}$. Taking $a=b$, we get $f(2a)=2f(a)$, so \[f(x+y)=f(x)+f(y),\ \forall x,y\in \mathbb{R}\] I have no clue how to continue Continue: Let $Q(x,y)$ be the assertion that $f(x^{3} + y^{3}) = x^{2}f(x) + y^{2}f(y).$ From $ Q(x-1,x+1) $ we get $f((x-1)^{3} + (x+1)^{3}) = f(2x^{3} + 6x) = (x^{2} - 2x + 1)*(f(x)-f(1)) + (x^{2} + 2x + 1)*(f(x)+f(1)) ==> f(2x^{3} + 6x) = 2f(x^{3}) + 2f(x) + 4xf(1).$On the other hand we know $f(2x^{3} + 6x) = 2*f(x^{3}) + 6f(x)$ Then we get $f(x) = x*f(1)$.

Sardor wrote: Find all functions $ f\colon \mathbb{R} \to \mathbb{R} $ such that $ f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy) $, for all $ x,y \in \mathbb{R} $. Let $P(x,y)$ be the assertion $f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy)$ $P(1,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x^3)=x^2f(x)$ $P(x,-x)$ $\implies$ $f(x)$ is odd Let then $u,v$ such that $u+v\le 0$. It's easy to see that the following system always has a solution : $x^3+y^3+xy=u$ $-x^3-y^3+xy=v$ $P(x,y)$ $\implies$ $f(u)=x^2f(x)+y^2f(y)+f(\frac{u+v}2)$ $P(-x,-y)$ $\implies$ $f(v)=-x^2f(x)-y^2f(y)+f(\frac{u+v}2)$ And so $f(\frac{u+v}2)=\frac{f(u)+f(v)}2$ $\forall u,v$ such that $u+v\le 0$ and so, since odd, $\forall u,v$ We immediately get from there that $f(x)$ is additive. Let then $p\in\mathbb N$ : $P(x+p,0)$ $\implies$ $f((x+p)^3)=(x+p)^2f(x+p)$ $\implies$ $p(2f(x)-2xf(1))=-3f(x^2)+2xf(x)+x^2f(1)$ $\forall p\in\mathbb N$ So $f(x)=xf(1)$ $\forall x$ And so $\boxed{f(x)=ax}$ $\forall x$, which indeed is a solution, whatever is $a\in\mathbb R$

pco wrote: Sardor wrote: And so $f(\frac{u+v}2)=\frac{f(u)+f(v)}2$ $\forall u,v$ such that $u+v\le 0$ and so, since odd, $\forall u,v$ We immediately get from there that $f(x)$ is additive. Can you explain this part?

Tima wrote: Can you explain this part? Just adding the two lines above gives $f(\frac{u+v}2)=\frac{f(u)+f(v)}2$ $\forall u,v$ such that $u+v\le 0$ If $u+v\ge 0$, we then have $-u-v\le 0$ and so $f(\frac{-u-v}2)=\frac{f(-u)+f(-v)}2$ and, since odd, $f(\frac{u+v}2)=\frac{f(u)+f(v)}2$ $\forall u,v$ such that $u+v\ge 0$ So $f(\frac{2u+0}2)=\frac{f(2u)+f(0)}2$ and so $f(2u)=2f(u)$ So $f(\frac{2u+2v}2)=\frac{f(2u)+f(2v)}2$ $=f(u)+f(v)$ and $f(x)$ is additive.

If we put $y=0$ in given equation we have $f(x^3)=x^2f(x)+f(0)$. Then if we put $x=1$ in the last one we have $f(0)=0$ and thus it becomes $f(x^3)=x^2f(x)$ or equivalently $f(x)=x^{\frac{2}{3}}f(x^{\frac{1}{3}})=x^{\frac{2}{3}+\frac{2}{9}}f(x^{\frac{1}{9}})=...=x^{\frac{2}{3}+\frac{2}{9}+...+\frac{2}{3^{n}}}f(x^{\frac{1}{3^{n}}})$ so if we let $n\rightarrow \infty$ we get $x^{\frac{2}{3}+\frac{2}{9}+...+\frac{2}{3^{n}}}=x$ and $f(x^{\frac{1}{3^{n}}})=f(1)$ and so the only solution is $f(x)=cx$ where $c$ is real.

bojler wrote: so if we let $n\rightarrow \infty$ we get $x^{\frac{2}{3}+\frac{2}{9}+...+\frac{2}{3^{n}}}=x$ and $f(x^{\frac{1}{3^{n}}})=f(1)$ Here you are assuming that $f$ is continuous which is not given as condition and has therefore first to be established (which I suppose won't be easy...).

can you tell me where the official solution is?

MGL wrote: can you tell me where the official solution is? http://matol.kz/comments/1537/show

Didn't know this problem was IZHO when I solved it in MOP.. Denote $P(x,y)$ as the assertion that $f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy)$. $P(1,0)$ gives $f(0)=0$. $P(x,0)$ gives $f(x^3)=x^2f(x)$. $P(x,-x)$ gives $f(x)=-f(-x)$. Adding $P(x,y)$ and $P(-x,-y)$ together and combine with the fact that $f$ is an odd function to get $$f(x^3+y^3+xy) + f(-x^3-y^3+xy)=2f(xy)$$ Fix $xy=c<0$. Then $x^3+y^3$ can is surjective in $\mathbb{R}$, so for $c<0$ we have $f(c+a)+f(c-a)=2f(c)$ for all $a$. By using that $f$ is odd, we have $f(a+b)+f(a-b)=2f(a)$ for all $a, b \in \mathbb{R}$, so we easily get that $f$ is additive. Now using everything we have, we get $$2x^2f(x)+6f(x)=f(2x^3+6x)=f((x+1)^3)+f((x-1)^3)=(x+1)^2(f(x)+f(1)) + (x-1)^2(f(x)-f(1)) = (2x^2+2)f(x) + 4xf(1)$$which gives $f(x)=f(1)x$, and we check that this works, so the answer is $\boxed{f(x) \equiv ax \text{ } \forall x \in \mathbb{R}}$

Thank for Sardor aka

rkm0959 wrote: Didn't know this problem was IZHO when I solved it in MOP.. Denote $P(x,y)$ as the assertion that $f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy)$. $P(1,0)$ gives $f(0)=0$. $P(x,0)$ gives $f(x^3)=x^2f(x)$. $P(x,-x)$ gives $f(x)=-f(-x)$. Adding $P(x,y)$ and $P(-x,-y)$ together and combine with the fact that $f$ is an odd function to get $$f(x^3+y^3+xy) + f(-x^3-y^3+xy)=2f(xy)$$ Fix $xy=c<0$. Then $x^3+y^3$ can is surjective in $\mathbb{R}$, so for $c<0$ we have $f(c+a)+f(c-a)=2f(c)$ for all $a$. By using that $f$ is odd, we have $f(a+b)+f(a-b)=2f(a)$ for all $a, b \in \mathbb{R}$, so we easily get that $f$ is additive. Now using everything we have, we get $$2x^2f(x)+6f(x)=f(2x^3+6x)=f((x+1)^3)+f((x-1)^3)=(x+1)^2(f(x)+f(1)) + (x-1)^2(f(x)-f(1)) = (2x^2+2)f(x) + 4xf(1)$$which gives $f(x)=f(1)x$, and we check that this works, so the answer is $\boxed{f(x) \equiv ax \text{ } \forall x \in \mathbb{R}}$ When xy is constant then x^3+y^3 is not surjective.

we have $P(x,y) = f(x^3 + y^3 + xy) = x^2f(x) + y^2f(y) + f(xy)$ $P(1,0) \implies f(0) = 0$ $P(x,0) : f(x^3) = x^2f(x) \implies f(x^3 + y^3 + xy) = f(x^3) + f(y^3) + f(xy)$ (1) $P(-x ,-y ) : f(-x^3 - y^3 + xy) = f(-x^3) + f(-y^3) + f(xy) = -f(x^3) - f(y^3) + f(xy)$ (2) (1) + (2) $\implies f(x^3 + y^3 + xy) + f(-x^3 - y^3 + xy) = 2f(xy)$ we get $a = x^3 + y^3 + xy and b = -x^3 - y^3 + xy$ $ \implies f(a) + f(b) = f(\frac{a+b}{2})$ (3) $ if xy \leq 0 \implies x^3 + y^3 can accept all \mathbb{R} , since f : odd$ To $ (3) : P(2a , 0) \implies f(2a) = 2f(a) \implies f(a) + f(b) = f(a+b) \implies f : additive $ To $ f(x^3) = x^2f(x) we give x \implies x + 1 (4) and x \implies x - 1 (5) then (4) + (5) $: $f(x) = x f(1)$ $O.Y.SH.$

Question $f({{x}^{3}}+{{y}^{3}}+xy)={{x}^{2}}f(x)+{{y}^{2}}f(y)+f(xy)$ $y=0\Rightarrow {{x}^{2}}f(x)=f({{x}^{3}})$ $x=0\Rightarrow {{y}^{2}}f(y)=f({{y}^{3}})$ So $f({{x}^{3}}+{{y}^{3}}+xy)=f({{x}^{3}})+f({{y}^{3}})+f(xy)$ i.e. we have $f(a+b+c)=f(a)+f(b)+f(c)$ so why not result that we have a Cauchy equation?

NO,He doesn,t say this, because only we say Cauchy equestion if $f(x+y)=f(x)+f(y)$