ZETA_in_olympiad wrote: Find all functions $f:\mathbb R\to \mathbb R$ such that \begin{align*} \left (x \left (f(x)-\dfrac{f(y)+f(z)}{2} \right) +y \left (f(y)-\dfrac{f(z)+f(x)}{2} \right ) +z\left (f(z)- \dfrac{f(x)+f(y)}{2} \right) \right )f(x+y+z)= \\ f(x^3)+f(y^3)+f(z^3)-3f(xyz) \end{align*}for all $x,y,z\in \mathbb R.$ Let $P(x,y,z)$ be the assertion $\frac 12(f(x)(2x-y-z)+f(y)(2y-x-z)+f(z)(2z-x-y))f(x+y+z)=f(x^3)+f(y^3)+f(z^3)-3f(xyz)$ Let $a=f(0)$, $c=f(1)$ and $d=f(-1)$ $P(1,0,0)$ $\implies$ $(c-1)(c-a)=0$ $P(-1,0,0)$ $\implies$ $(d+1)(d-a)=0$ $P(1,1,-1)$ $\implies$ $(c-1)(c-d)=0$ $P(1,-1,-1)$ $\implies$ $(d+1)(c-d)=0$ And so either $c=1$ and $d=-1$, either $c=d=a$ 1) If $c=d=a$ $P(x,1,-1)$ $\implies$ $x(f(x)-a)f(x)=f(x^3)+2a-3f(-x)$ $P(x,0,0)$ $\implies$ $f(x^3)=xf(x)^2-axf(x)+a$ Adding, we get $f(-x)=a$ and so $\boxed{\text{S1 : }f(x)=a\quad\forall x}$, which indeed fits, whatever is $a\in\mathbb R$ 2) If $c=1$ and $d=-1$ $P(0,1,-1)$ $\implies$ $a=0$ $P(x,0,0)$ $\implies$ $f(x^3)=xf(x)^2$ $P(x,1,-1)$ $\implies$ $f(-x)=-f(x)$ $P(x,y,-y)$ $\implies$ $f(xy^2)=yf(x)f(y)$ Setting there $x=1$, we get $f(y^2)=yf(y)$ and so $f(xy^2)=f(x)f(y^2)$ So $f(xy)=f(x)f(y)$ $\forall y\ge 0$ and so, since odd, $\forall x,y$ So $f(x^3)=f(x)^3$ and so $f(x)^3=xf(x)^2$ So $\forall x$, either $f(x)=0$, either $f(x)=x$ If $f(u)=0$ for some $u\ne 0$, $1=f(1)=f(u\frac 1u)=f(u)f(\frac 1u)=0$, contradiction And so $\boxed{\text{S2 : }f(x)=x\quad\forall x}$, which indeed fits.

Same solution as #2, posting for storage. Let $P(x,y,z)$ denote the given assertion. We claim that the only solution are constant functions or identity functions, both clearly work. $P(0,0,1)\implies f(1)-f(0)=f(1)(f(1)-f(0))$ $P(0,0,-1)\implies f(-1)-f(0)=f(-1)(f(0)-f(-1))$ $P(0,1,-1)\implies 2(f(1)+f(-1))=f(0)(3f(1)-3f(-1)+4)$ $P(1,1,-1)\implies 2(f(1)-f(-1))=2f(1)(f(1)-f(-1))$ $P(1,-1,-1)\implies 2(f(-1)-f(1))=2f(-1)(f(1)-f(-1))$ From above we deduce, $(f(0),f(1),f(-1))=(0,1,-1)$ or $f(0)=f(1)=f(-1)=k.$ Considering the latter, comparing $P(x,0,0)$ and $P(x,1,-1)$ we get $f\equiv k.$ Consider $(f(0),f(1),f(-1))=(0,1,-1)$, then $P(x,0,0)$ gives $f(x^3)=xf(x)^2.$ And $P(x,1,-1)$ implies $f$ is odd. So $P(x,-x,1)$ implies $f(x^2)=xf(x).$ And so, with oddness, $P(x,-x,y)$ implies $f$ is multiplicative. In particular $f(x_0)=0$ for $x_0\neq 0$ implies $f\equiv 0,$ false. So $xf(x)=f(x^2)=f(x)^2$ means $f(x)=x.$