Find all polynomials $P(x) \in \mathbb{R}[x]$ such that $$P(xy)+P(yz)+P(zx) =P(xy+yz+zx)$$
$\forall x, y, z \in \mathbb{R}$ satisfying the equation $x+y+z=0$
My attempt :
Consider $P(0,0,0)$, $3P(0)=P(0)$, so $P(0)=0$
Since $x+y+z=0$, let $x=a-b, \;y=b-c, \;z=c-a, \;\forall a, b, c \in \mathbb{R}$
$P((a-b)(b-c)) + P((b-c)(c-a)) + P((c-a)(a-b))$
$= P((a-b)(b-c)+(b-c)(c-a)+(c-a)(a-b))$
$= P((a-b)(b-c)+(c-a)(a-c))$
$= P(ab+bc+ca-a^2-b^2-c^2)$
Put $y=1$, $z=-x-1$. This gives $$P(-x^2-x-1)-P(-x^2-x)=P(x)+P(-x-1)$$ Now suppose that $P(u)=a_mu^m+\cdots$. There exists a $c\in ]-x^2-x-1,-x^2-x[$ such that $$P(-x^2-x-1)-P(-x^2-x)=-P^{\prime}(c)$$ If $x\to +\infty$, we get $P^{\prime}(c)\sim ma_m (-1)^{m-1} x^{2m-2}$. If $m$ is even, $P(x)+P(-x-1)\sim 2a_m x^m$. This show that if $m$ is even, we have $m=2$.
If $m$ is odd, then $|P(x)+P(-x-1)|\leq cx^{m-1}$, for some constant $c>0$, and for large $x$. This show that $2m-2\leq m-1$, ie $m\leq 1$.
Now, noticing that the polynomials of degree $1$ (such that $P(0)=0$) works, and that a linear combination of solutions is again a solution, it remains to see if the polynomial $P(u)=u^2$ work or not, I leave it to you.
Working first with $y=x$ and $z=-2x$, we obtain $$P(x^2)+ 2 P(-2 x^2) = P(-3 x^2)$$ Let $a_n x^n$ be the highest order term in $P(x)$, this relation implies $$a_n x^{2n} + 2 (-2)^n a_n x^{2n} = (-3)^n a_n x^{2n} \quad\Longrightarrow\quad 1 + 2 (-2)^n = (-3)^n$$ The only solutions are $n = 1$ or $n = 2$. So that $P$ has degree $1$ or $2$. Observe also that $P(0) = 0$ so that there is no constant term in $P$ and also that the set of solutions is a vector space. It only remains to check that the polynomials $x$ and $x^2$ satisfy the initial property, finally $$P(x) = \alpha x^2 + \beta x \qquad \alpha, \beta \in {\mathbb R}$$
