Problem 1386

1. Substitution and Simplification:
We start by substituting $x = a - b$ and $y = b - c$ into the given equation:
$$p(a + b - 2c) + p(b + c - 2a) + p(c + a - 2b) = 3p(a - b) + 3p(b - c) + 3p(c - a)$$
This transforms the equation into:
$$p(x - y) + p(-2x - y) + p(x + 2y) = 3p(x) + 3p(y) + 3p(-x - y)$$

2. **Special Case $x = y = 0$:**
Setting $x = 0$ and $y = 0$ in the transformed equation, we get:
$$p(0) + p(0) + p(0) = 3p(0) + 3p(0) + 3p(0)$$
Simplifying, we find:
$$3p(0) = 9p(0) \implies p(0) = 0$$

3. **Special Case $y = 0$:**
Setting $y = 0$ in the transformed equation, we get:
$$p(x) + p(-2x) + p(x) = 3p(x) + 3p(-x)$$
Simplifying, we find:
$$2p(x) + p(-2x) = 3p(x) + 3p(-x) \implies p(-2x) = p(x) + 3p(-x)$$

4. Assuming Polynomial Form:
Assume $p(x)$ is a polynomial of degree $n$:
$$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$
Since $p(0) = 0$, we have $a_0 = 0$. Thus:
$$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x$$

5. **Analyzing the Equation $p(-2x) = p(x) + 3p(-x)$:**
Substitute $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x$ into the equation:
$$a_n (-2x)^n + a_{n-1} (-2x)^{n-1} + \cdots + a_1 (-2x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + 3(a_n (-x)^n + a_{n-1} (-x)^{n-1} + \cdots + a_1 (-x))$$
Simplifying, we get:
$$a_n (-2)^n x^n + a_{n-1} (-2)^{n-1} x^{n-1} + \cdots + a_1 (-2) x = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + 3(a_n (-1)^n x^n + a_{n-1} (-1)^{n-1} x^{n-1} + \cdots + a_1 (-1) x)$$

6. Equating Coefficients:
For each power of $x$, we equate the coefficients:
$$a_k (-2)^k = a_k + 3a_k (-1)^k$$
This simplifies to:
$$a_k \left[ (-2)^k - 1 - 3(-1)^k \right] = 0$$
This gives us two cases:
$$\begin{cases} a_k (2^k - 4) = 0 & \text{if } k \text{ is even} \\ a_k (2^k - 2) = 0 & \text{if } k \text{ is odd} \end{cases}$$

7. Solving for Coefficients:
For $k \geq 3$, both equations yield $a_k = 0$. Therefore, the polynomial $p(x)$ must be of degree at most 2:
$$p(x) = \lambda x^2 + \mu x$$
where $\lambda$ and $\mu$ are real constants.

8. Verification:
Substituting $p(x) = \lambda x^2 + \mu x$ back into the original equation, we find that it satisfies the equation for any $\lambda, \mu \in \mathbb{R}$.

The final answer is $\boxed{ p(x) = \lambda x^2 + \mu x }$, where $\lambda, \mu \in \mathbb{R}$.