$p(1,0) \implies f(1)=c=+1$ or $-1$ $\implies f(x)=xc, \quad \forall x \in (+\frac{1}{4},+\infty)$ $f(a).f(x^2+(a-x)^2-(a-x)x)=a(x^2+(a-x)^2-(a-x)x)$ such as that $a \le \frac{1}{4}$ now because $x^2+(a-x)^2-(a-x)x$ is quadratic polynomial in terms of $x$ with the coefficient of $x^2$ positive then for any $a$ we know that the polynomial will be more that $\frac{1}{4}$ after some certain point thereupon by putting $x_0$ such a values it implies that $f((x_0)^2+(a-x_0)^2-(a-x_0)x_0)=(x_0)^2+(a-x_0)^2-(a-x_0)x_0 \implies f(a)=ac$ for constant $c$ which can equal to $-1,+1$

Let $P(x,y):=f(x+y)f(x^2-xy+y^2)=x^3+y^3$ $P(x,0):f(x)f(x^2)=x^3$ So $f(1)= \mp 1$ Consider that if $f$ satisfies the condition, $-f$ do too. So assume $f(1)=1$ Also $P(x,0),P(-x,0): f(-x)=-f(x)$ $P(x,1-x):f(1)f(3x^2-3x+1)=3x^2-3x+1$ Let $y=3x^2-3x+1$ $\frac{d}{dx}3x^2-3x+1=6x-3$ So minimum of $y$ is equal to $\frac{1}{4}$ when $x=\frac{1}{2}$ Now we know $f(x)=x, \quad \forall x \in [-\infty,-\frac{1}{4}) \cup (+\frac{1}{4},+\infty]$ But how about $|x|<\frac{1}{4}$? See... Imagine for some numbers like one $q$ we have $f(q)\ne q$ Using $P(x,0):f(x)f(x^2)=x^3$ we can grow $x$ til it get bigger that $\frac{1}{4}$ For example for $f(\frac{1}{16}$ we have $f(\frac{1}{4})f(\frac{1}{16})=\frac{1}{4}\times \frac{1}{16}$ And now more general for every $t>0$ smaller than $\frac{1}{4}$ we can use the relation that is divergent on $<1$ or use conjunction and $\sqrt{t}>t$ And it gives contradiction with our assumption $f(q)\ne q$ (Recursive relationship $f(\sqrt{q})=\frac{q\sqrt{q}}{f{q}}$) and we are done so only solutions are $\boxed {f(x)=\mp x}$

Let $t=x+y$ and $r=x^2+y^2-xy$. I claim that for $0<t<4$ there exist $x,y\in \mathbb{R}$ such that $t=r$: $y=t-x\Longrightarrow t^2-3x(t-x)=t\Longrightarrow 3x^2-3xt+(t^2-t)=0$. But it's obvious for $4\geq t\geq 0$ we have $\triangle=-3t^2+12t\geq 0$ so it has got real roots. So there exist $r,t$ such that $r=t\Longrightarrow f(t)^2=t^2\Longrightarrow f(t)=\pm t$. choose $4\geq t\geq 0$ and $y=t-x\Longrightarrow f(3x^2+t^2-3tx)=\pm(3x^2+t^2-3tx)$ but $\text{min}\{3x^2+t^2-3tx\}=\dfrac{t^2}{4}$ hence $3x^2+t^2-3tx$ covers all nonegative real numbers. So for $x>0$: $f(x)=\pm x$. Finally since $x^2+y^2-xy=(x-\frac{y}{2})^2+\frac{3}{4}y^2\geq 0$ we get for every $x,y$ , $f(x+y)=\pm(x+y)\Longrightarrow f(x)=\pm x$ for $x\in \mathbb{R}$. obviously seen that $f(x)=x$ or $f(x)=-x$ for $x\in \mathbb{R}$.

$P(x,y)=f(x+y)f(x^2-xy+y^2)=x^3+y^3$

First note that if $f(x^2+y^2-xy)=0$ then by plugging in the euation we get $x+y=0$.Now let $g(x)=\frac{f(x)}{x} \forall x \neq 0$ Then we can rewrite the equation as: $g(x+y)=\frac{1}{g(x^2+y^2-xy)} \forall x \neq y$ One can easily check that the equation $x+y=a,x^2+y^2-xy=b$ has a solution if and only if $b \ge \frac{a^2}{4}$ So we can rewrite the equation as: $P(a,b):g(a)=\frac{1}{g(b)} \forall b \ge \frac{a^2}{4}$ Now lets take $x_0 \neq 0$ I will prove $g(x_0)=g(1)$.Lets take $y_0$ big enough so that $y_0 \ge \max \{\frac{1}{4},\frac{x_0^2}{4} \}$. $P(1,y_0),P(x_0,y_0) \Rightarrow g(x_0)=g(1)$ Since $x_0$ was arbitary We have $g(x)=c \forall x \neq 0 \Rightarrow f(x)=cx \forall x \neq 0$.By plugging $x=y=0$ in the original equation we get $f(0)=0$ So $f(x)=cx \forall x \in \mathbb{R}$.By plugging in the original equation we get $f(x)=x \forall x \in \mathbb{R} ,f(x)=-x \forall x \in \mathbb{R}$ are the only solutions.

its my solution:

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A short solution: Let $h(x,y)=x^2-xy+y^2$ Easy to see that $h(x,y)=h(x,x-y)$ Thus we have: $\frac{f(x+y)}{f(2x-y)}=\frac{x+y}{2x-y}$ And now $y\to 2x-1$: $f(3x-1)=(3x-1)f(1)$ And same as before $f(1)=\pm 1$ thus: $f(x)=\pm x \forall x$

Etemadi wrote: Find all functions $f:\Bbb {R} \rightarrow \Bbb {R} $ such that: $$f(x+y)f(x^2-xy+y^2)=x^3+y^3$$for all reals $x, y $. I'm not quite a bid sure about my solution: So by setting $P(x,y)=(0,0)$ we get $f(0)=0$ and $P(x,y)=(1,0)$ then $f(1)=\pm 1$ Then by setting $P(x,y)=(x,0)$ we get: $f(x)f(x^2)=x^3$ We can also set $P(x,y)=(x,x)$: $f(2x)f(x^2)=2x^3$ Or $\frac{f(2x)}{2}=f(x)$ Now we can guess the function is constant as $f(x)=ax+$c, hence: $\frac{2ax+c}{2}=ax+c$ Or $c=0$ Then the function becomes $f(x)=ax$ and when we replace it to the original function we get: $a(x+y)a(x^2-xy+y^2)=x^3+y^3$ Or $a^2=1$ that means $a=\pm 1$ hence,$f(x)=\pm x$ So we're done...!

iliyash wrote: Now we can guess the function is constant as $f(x)=ax+$ce It should be explained... You should check $P(x,1-x)$ then: $f(3x^2-3x+1)=3x^3-2x+1$ So $f(x)=x$ for all $|x|$ greater than $\frac{1}{4}$ Now can use your formula to say that $f(x)=x$ for all $|x|$ smaller than $\frac{1}{4}$. Such a way: $f(t)=\frac{f(2^nt)}{2^n}$

$P(x,y)=f(x+y)f(x^2-xy+y^2)=x^3+y^3$ $P(0,0)=f(0)f(0)=0 \Rightarrow f(0)=0$ $P(1,0)=f(1)f(1)=1\Rightarrow $$f(1)=\pm 1$ if $f(1)=1 \Rightarrow g(x)=\frac{f(x)}{x}$ , $ g:\Bbb {R}-{0} \rightarrow \Bbb {R} $ $g(1)=1$ $g(x+y)(x+y)g(x^2-xy+y^2)(x^2-xy+y^2)=x^3+y^3$ $(x^2-xy+y^2)(x+y)=x^3+y^3$ so we have $Q(x,y)=g(x+y)g(x^2-xy+y^2)=1$ $Q(x,1-x)=g(1)g(3x^2-3x+1)=1\Rightarrow g(3x^2-3x+1)=1 $ if $ z=3x^2-3x+1 \geq 1 $have root because $\triangle=9-12(1-z)\geq 0$ so $g(3x^2-3x+1)$ in $\quad \forall x \in [1,+\infty]$ is surjective so suppose that $\exists_s \Rightarrow g(s) \neq s$ $Q(x,s-x)=g(ss)g(x^2-x(s-x)+(s-x)^2)=1\Rightarrow g(3x^2-3xs+s^2) \neq 1 $ if $s^2 \le z=(3x^2-3xs+s^2)$ so $z=(3x^2-3xs+s^2)$ have a root because $\triangle=9s^2-12(s^2-z)\geq 0$ so $\forall_s s^2\le z \Rightarrow g(z)\neq 1 $ suppose that $z \geq 1+s^2$ on the other hand , we know $g(z)=1$ on the other hand , we know $g(z)\neq 1 \Rightarrow $ so $\nexists_s , z \geq 1+s^2$ so $\forall_x , x \geq 0 \Rightarrow g(x)=1 \Rightarrow f(x)=x$ and $f(0)=0 \Rightarrow \forall_x , f(x)=x$ $(x+y)(x^2-xy+y^2)=x^3+y^3$ similarly if $f(1)=-1 \Rightarrow g(x)=\frac{f(x)}{x}$ , $ g:\Bbb {R}-{0} \rightarrow \Bbb {R} $ $g(x)=-1 , \forall_x ,x \geq 0 \Rightarrow f(x)=-x $ $f(0)=0$ $\Rightarrow f(x)=-x , \forall_x$ $(x+y)(x^2-xy+y^2)(-1)^2=x^3+y^3$ so we are done so only solutions are $f(x)=\pm x$

This is my solution

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Note that equation $x+y = x^2 - xy + y^2$ has real roots cause $x+y = (x+y)^2 - 3x((x+y)-x) \implies 3x^2-3x(x+y)+((x+y)^2-(x+y))=0$ and for $4\ge (x+y)\ge 0$ we have $\triangle \ge 0$. Let $a = (x+y) = x^2 - xy + y^2$ so $f(a)^2 = (x+y)(x^2 - xy + y^2) = a^2 \implies f(a) = \pm a$. Now we choose $4\ge (x+y)\ge 0$ and then $f(3x^2-3x(x+y)+((x+y)^2-(x+y))) = \pm 3x^2-3x(x+y)+((x+y)^2-(x+y))$ and we know $3x^2-3x(x+y)+((x+y)^2-(x+y))$ covers all nonegative real numbers so $x \ge 0: f(x)=\pm x$. $P(x,0) : f(x)f(x^2) = x^3$ $P(-x,0) : f(-x)f(x^2) = -x^3$ $\implies f(x) = -f(-x)$. so now we have $f(x)=x$ or $f(x)=-x$. Answers : $f(x)=x , f(x)=-x$

The answers are $f(x) = x$ for all $x\in \mathbb{R}$ and $f(x)=-x$ for all $x \in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(x+y)f(x^2-xy+y^2)=x^3+y^3$ . We start off by making some preliminary observations. Claim : For any such function $f$ , $f(x)=0$ if and only if $x=0$. Proof : We can note that from $P(0,0)$ we have that $f(0)^2=0$ so $f(0)=0$ quite clearly. Now, say there exists some $\alpha \ne 0 \in \mathbb{R}$ such that $f(\alpha)=0$. Then, $P(\alpha,0)$ yields, \[0=f(\alpha)f(\alpha^2)=\alpha^3 \ne 0\]which is a clear contradiction. Now, we can prove some properties of $f$ with these observations in hand. Claim : For all real numbers $x \ne 0$ , $f(-x)=-f(x)$ and $f(2x)=2f(x)$. Proof : From $P(-x,-y)$ we have that, \[f(-(x+y))f(x^2-xy+y^2)=-(x^3+y^3)=-f(x+y)f(x^2-xy+y^2)\]Considering $x+y=t$ such that $x^2-xy+y^2 \ne 0$ yields that, $f(-t)=-f(t)$ for all $t\in \mathbb{R}$ as desired. Next, from $P(x,x)$ we have, \[f(2x)f(x^2)=2x^3\]and from $P(x,0)$ we have, \[f(x)f(x^2)=x^3\]Thus, \[2f(x)f(x^2)=2x^3 = f(2x)f(x^2)\]which implies that $f(2x)=2f(x)$ for all $x\ne 0$, as claimed. Now, we are almost done. From $P(1,1-x)$ we have that \[f(1)f(3x^2-3x+1)=3x^2-3x+1\]It is easy to check that the term $3x^2-3x+1$ is surjective over $[\frac{1}{4},\infty)$ and thus, $f(t)=ct$ for all $t\ge \frac{1}{4}$ where $c$ is a constant ($c= \frac{1}{f(1)}$ , note that $f(1)\ne 0$ as a result of our first claim). Since $f(2x)=2f(x)$ by simple induction we have that $f(2^rx)=2^rf(x)$ for all positive integers $r$. Thus, for all positive reals $x$, \[2^rf(x)=f(2^rx)=2^rcx\]so $f(x)=cx$ for all positive real numbers $x$ (by considering sufficiently large $r$), which then combined with the fact that $f$ is odd, allows us to conclude that $f(x)=cx$ for all real numbers $x$. Now, all that remains is to substitute the obtained expression into the given equation and check which $c$ work. A short calculation yields $c^2=1$ so $c\in \{1,-1\}$ of which both can be seen to work.