Define
$$Q(x, y, z) = xP(x) + yP(y) + zP(z) - xyz[P(x - y) + P(y - z) + P(z - x)]$$
Then $Q(x, y, z)$ is also a polynomial with real coefficients, and when $xyz \neq 0$,
$$2xyz = x + y + z \Rightarrow Q(x, y, z) = 0$$
The property above can be extended to complex numbers, that is, $x, y, z$ can also be substituted with complex numbers. Substituting $(x, y, z) = (t, -t, 0)$ gives $P(t) = P(-t)$, so $P(x)$ is an even function. Substituting further
$$(x, y, z) = \left(x, \frac{i}{\sqrt{2}}, -\frac{i}{\sqrt{2}}\right)$$
we obtain
$$\begin{aligned} & xP(x) + \frac{i}{\sqrt{2}}\left(P\left(\frac{i}{\sqrt{2}}\right) - P\left(-\frac{i}{\sqrt{2}}\right)\right) \\ & = \frac{1}{2}x\left(P\left(x - \frac{i}{\sqrt{2}}\right) + P\left(x + \frac{i}{\sqrt{2}}\right) + P(\sqrt{2}i)\right) \end{aligned}$$
which leads to
$$P\left(x + \frac{i}{\sqrt{2}}\right) + P\left(x - \frac{i}{\sqrt{2}}\right) - 2P(x) = P(\sqrt{2}i)$$
This shows that $\deg P(x) \le 2$, and the general form of $P(x)$ is $ax^2 + b$.