. Find all polynomials $p(x)$ with real coefficients such that

$$
p(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)
$$

for all $a, b, c \in \mathbb{R}$.
. For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have

$$
p(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)
$$

for all $a \in \mathbb{R}$. So we find a polynomial equation

$$
p(-2 x)=p(x)+3 p(-x) \text {. }
$$

Note that the zero polynomial is a solution to this equation. Now suppose that $p$ is not the zero polynomial, and let $n \geq 0$ be the degree of $p$. Let $a_{n} \neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (1), the coefficient of $x^{n}$ is $(-2)^{n} \cdot a_{n}$, while at the right-hand side the coefficient of $x^{n}$ is $a_{n}+3 \cdot(-1)^{n} \cdot a_{n}$. Hence $(-2)^{n}=1+3 \cdot(-1)^{n}$. For $n$ even, we find $2^{n}=4$, so $n=2$, and for $n$ odd, we find $-2^{n}=-2$, so $n=1$. As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).

The polynomial $p(x)=x$ is a solution to our problem, as

$$
(a+b-2 c)+(b+c-2 a)+(c+a-2 b)=0=3(a-b)+3(b-c)+3(c-a)
$$

for all $a, b, c \in \mathbb{R}$. Also, $p(x)=x^{2}$ is a solution, since

$$
\begin{aligned}
(a+b-2 c)^{2}+(b+c-2 a)^{2}+(c+a-2 b)^{2} & =6\left(a^{2}+b^{2}+c^{2}\right)-6(a b+b c+c a) \\
& =3(a-b)^{2}+3(b-c)^{2}+3(c-a)^{2}
\end{aligned}
$$

for all $a, b, c \in \mathbb{R}$.

Now note that if $p(x)$ is a solution to our problem, then so is $\lambda p(x)$ for all $\lambda \in \mathbb{R}$. Also, if $p(x)$ and $q(x)$ are both solutions, then so is $p(x)+q(x)$. We conclude that for all real numbers $a_{2}$ and $a_{1}$ the polynomial $a_{2} x^{2}+a_{1} x$ is a solution. Since we have already shown that there can be no other solutions, these are the only solutions.