Here is the initial piece of the solution: (1). Set $x=y=0$ to deduce $f(0)=0$. (2). Set $y=0$ to deduce $f(x^3)=xf(x^2)$. (3). By using (2), the given equation simplifies to $xf(y^2)+yf(x^2)=(x+y)f(xy)$. (4). For negative $x$, let $x=-a^3$. Then (2) yields $f(x)=f(-a^3)=-af(a^2)=-f(a^3)=-f(-x)$. Hence the function is skew-symmetric and satisfies $f(-x)=-f(x)$ for all $x$. From now on we concentrate on positive real numbers. (5). Set $y=x^2$ in (2) to deduce $xf(x^4)+x^2f(x^2)=(x+x^2)f(x^3)=(x+x^2)xf(x^2)$. Hence $f(x^4)=x^2f(x^2)$ for all $x\ne0$. (6). For positive $x$, let $x=a^2$. Then (5) yields $f(x^2)=xf(x)$ for all $x\ge0$. (7). By using (6), the equation (2) simplifies to $xy(f(x)+f(y))=(x+y)f(xy)$.

Here is the remaining piece of the solution: (8). Set $y=1$ in (7) to deduce $xf(1)=f(x)$. Hence, for $x\ge0$ the function is linear and of the form $f(x)=cx$ for some real $c$. From (4) we see that also for $x<0$ we have $f(x)=cx$. (9). By plugging $f(x)=cx$ into the original equation, we see that all linear functions indeed are solutions.

dorina wrote: Solve the function $f: \Re \to \Re$ $ f( x^{3} )+ f(y^{3}) = (x+y)(f(x^{2} )+f(y^{2} )-f($x*y$ )) $ Leje Dorin se po i postoj un te gjitha

As obtained by others $f(x^3) = x f(x^2)$. From $y \rightarrow -x$ we see that $f(-t) = -f(t)$ Now setting $y \rightarrow -y$ we get $f(x^3)-f(y^3) = (x-y)(f(x^2)+f(y^2)+f(xy))$ Adding this to the given functional equation we obtain $f(x^3) = xf(x^2) + xf(y^2)-yf(xy) \Rightarrow xf(y^2) = yf(xy)$ Setting $y=1$ its immediate that $f(x) = xf(1)$. So $f(x) = cx$ is the possible set of solutions. Checking we see that $c=1,0$ are the only admissible values of c. Hence $f(x)=x$ or $f(x)=x \ \forall x \in \mathbb{R}$ are the only solutions

How did you check?!!! I think f(x)=cx is perfectly right and satisfies the conditions...

dorina wrote: Solve the function $f: \Re \to \Re$ $ f( x^{3} )+ f(y^{3}) = (x+y)(f(x^{2} )+f(y^{2} )-f($x*y$ )) $ Guys, what happened to you? Checking shows $f(x)=cx$ for any real $c$ is a solution.

[Edit: To fix a typo] @LeMagnifiqueEp, both you and I know that I meant $f(x)=-f(-x)$, but accidentally did a typo.

f(x)=-f(x) is obviously not true! that means f(x)=0.... check your work (by the way, your way is obviously correct!)

Quite Easy to get $f(x^3)=xf(x^2)$ and $f(-x)=-f(x)$ , $\forall x \in \mathbb{R}$ ... Now using $f(x^3)=xf(x^2)$, the original equation becomes $xf(y^2)+yf(x^2)=(x+y)f(xy)$. Plug $y=1$ and $y=-1$, add them and get $f(x)=\alpha x$ , $\forall x \in \mathbb{R}$. [where $\alpha = f(1)$, a constant]

This is also British MO 2008/09

i've seen this problem in two other competitions too

Solution 1) $x=y=0$ $=f(0^3)+f(0^3)=(0+0)f(0^2)+f(y^2)–f(0.0)$ $=f(0)=f(0)+f(y^2)–f(0)$ $=f(0)=\frac{f(0^2)}{2}$ $f(0)=0$ 2nd $x=0$ $f(y^3)=f(y^2)$ 3rd $y=0$ $f(x^3)=xf(x^2)$ From above it dollows $f(x^2)=xf(x^2)$ $x=0$ (impossible) $f(x)=0$

dorina wrote: Solve the function $f: \Re \to \Re$ $ f( x^{3} )+ f(y^{3}) = (x+y)(f(x^{2} )+f(y^{2} )-f($x*y$ )) $ Can you give more information? about year? Albania IMO Selection??

Pham_Quoc_Sang wrote: Can you give more information? about year? Albania IMO Selection?? Albania TST 2013 Dorina wrote: Solve the function $f: \Re \to \Re$: \[ f( x^{3} )+ f(y^{3}) = (x+y)(f(x^{2} )+f(y^{2} )-f(xy))\] Nice FE problem. SOLUTION CLAIM 1: $f(0)=0$ Proof. Let $x=y=0$ to yield that $f(0)+f(0)=0(f(0)+f(0)-f(0)) \implies f(0)=0$. CLAIM 2: $f(x)=-f(-x)$ Proof. Let $y=-x$. It follows that $f(x^3)+f(-x^3)=(x-x)(f(x^2)+f(x^2)-f(-x^2)) \implies$ $f(x^3)+f(-x^3)=0 \implies$ $f(x)=-f(-x)$. CLAIM 3: $f(x^3)=xf(x^2)$ Proof. Let $y=0$. It follows that $f(0)+f(x^3)=(y+0)(f(0)+f(x^2)-f(0)) \implies$ $f(x^3)=xf(x^2)$. CLAIM 4: $f(x)=xf(1)$ Proof. Applying claim $3$ to the original equation, we have that $xf(x^2)+yf(y^2)=(x+y)((f(x^2)+f(y^2)-f(xy)) \implies$ $(x+y)f(xy)=yf(x^2)+xf(y^2)$. Letting $y=1$ and $y=-1$ respectively, we have that $$(x+1)f(x)=f(x^2)+xf(1)$$$$(x-1)f(-x)=-f(x^2)+xf(1)$$Adding the two equations, we have that $f(x)-f(-x)+x(f(x)+f(-x))=2xf(1) \implies$ $f(x)-f(-x)=2xf(1) \implies$ $f(x)=xf(1)$. CLAIM 5: $f(x)=cx$ Proof. Replacing $f(x)$ with $cx$ in the original equation yields: $(cx)^3+(cy)^3=(x+y)((cx)^2+(cy)^2-(cx)(cy))$. Expanding the LHS, we see that this is indeed true.

I have discussed this problem on my YouTube channel. Video Solution

$P(0,0): 2f(0)=0\implies f(0)=0$. $P(x,-x): f(x^3)+f(-x^3)=0$, which implies $f$ is odd. $P(x,0): f(x^3)=xf(x^2)$. So $xf(x^2)+yf(y^2)=xf(x^2)+yf(y^2)+xf(y^2)+yf(x^2)-xf(xy)-yf(xy)\implies xf(y^2)+yf(x^2)=(x+y)f(xy)$. Let $Q(x,y)$ denote the new assertion. Since $f$ is odd, we can only focus on positive reals as $f(0)=0$. $Q(x,x^2): xf(x^4)+x^2f(x^2)=(x+x^2)f(x^3)=xf(x^3)+x^2f(x^3)=x^2f(x^2)+x^3f(x^2)\implies xf(x^4)=x^3f(x^2)\implies f(x^4)=x^2f(x^2)$. So now we know that $f(x^2)=xf(x)$. Thus, the new assertion $R(x,y)$ becomes $xyf(x)+xyf(y)=xf(xy)+yf(xy)$ $R(1,x): xf(1)+xf(x)=f(x)+xf(x)\implies f(x)=xf(1)$. So the only solution is $\boxed{f(x)=cx\text{ for all }x\in\mathbb{R} \text{ and all constants }c\in\mathbb{R}}$, which work.

LeMagnifiqueEp wrote: f(x)=-f(x) is obviously not true! that means f(x)=0.... check your work (by the way, your way is obviously correct!) 事实上他应该少了一个负号。。。

@above what are you trying to convey? ---- Either I'm forgetting things or I I've done this FE before.

$P(x,-x)\implies f(x^3)+f(-x^3)=0 \implies f \text{ is odd } \implies f(0)=0.$ $P(x,0)\implies f(x^3)=xf(x^2).$ The original equation becomes, $(x+y)f(xy)=xf(y^2)+yf(x^2).$ $P(x,-y) \implies (y-x)f(xy)=xf(y^2)-yf(x^2).$ Add the two to get, $yf(xy)=xf(y^2).$ $P(x,1) \implies f(x)=xf(1),$ set $f(1)=a \implies \boxed{f(x)=ax},$ which fits.

dorina wrote: Solve the function $f: \Re \to \Re$: \[ f( x^{3} )+ f(y^{3}) = (x+y)(f(x^{2} )+f(y^{2} )-f(xy))\] $$f( x^{3} )+ f(y^{3}) = (x+y)(f(x^{2} )+f(y^{2} )-f(xy))...(\alpha)$$In $(\alpha) x=y=0:$ $$\Rightarrow f(0)=0$$In $(\alpha) y=0:$ $$\Rightarrow f(x^3)=xf(x^2)...(\beta)$$$$\Rightarrow f \text{ is odd}...(I)$$$(\beta)$ in $(\alpha):$ $$\Rightarrow xf(x^2)+yf(y^2)=xf(x^2)+xf(y^2)-xf(xy)+yf(x^2)+yf(y^2)-yf(xy)$$$$\Rightarrow xf(xy)+yf(xy)=xf(y^2)+yf(x^2)...(\theta)$$In $(\theta) y=-y:$ $$\Rightarrow -xf(-xy)+yf(-xy)=-xf(y^2)+yf(x^2)$$By $(I):$ $$\Rightarrow xf(xy)-yf(xy)=-xf(y^2)+yf(x^2)...(\omega)$$$(\omega)+(\theta):$ $$\Rightarrow 2xf(xy)=2yf(x^2)...(\phi)$$In $(\phi) x=1:$ $$\Rightarrow f(y)=yf(1), \forall y\in \mathbb{R}-\{ 0\}$$ $$\Rightarrow f(x)=cx, c \text{ is constant}, \forall x\in\mathbb{R}_\blacksquare$$

Hang on a minute ,if you assume that $x=y$,then you get $f(x^{3})=xf(x^{2})$ . Then you can take this to the assumption, and you get $yf(x^{2})+xf(y^{2})=(x+y)f(xy)$ is true for any real $x,y$. Let y transfers to -y in the former function , we get $-yf(x^{2})+xf(y^{2})=(y-x)f(xy)$,(it's easy to get $f(x)=-f(-x)$ then we add two functions. The answer goes to $xf(y^{2})=yf(xy)$, let y=1,then we get $xf(1)=f(x)$. Hence $f(x)=cx$ (in which c is a constant) (It's not necessary to get $f(0)=0$)