<s>[INST] Polynomials : $P(xy)+P(yz)+P(zx) =P(xy+yz+zx)$ 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 $x=a-b, \;y=b-c, \;z=c-a, \;\forall a, b, c \in \mathbb{R}$
$P((a-b)(b-c)) + P((b-c)(c-a)) + P((c-a)(a-b))$
$= P((a-b)(b-c)+(b-c)(c-a)+(c-a)(a-b))$
$= P((a-b)(b-c)+(c-a)(a-c))$
$= P(ab+bc+ca-a^2-b^2-c^2)$
 [/INST] 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
$$P(x) = \alpha x^2 + \beta x \qquad \alpha, \beta \in {\mathbb R}$$
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