Let $P(x)$ satisfy the given equation.
$$\text{If } a = b = c = 0, \text{ then } P(0) = 0.$$
$$\text{If } b = c = 0, \text{ then } P(-a) = P(a) \text{ for all real } a.$$
Hence $P(x)$ is even. Without loss of generality, we may assume that
$$P(x) = a_n x^{2n} + \cdots + a_1 x^2,\ a_n \neq 0.$$
If $b = 2a,\ c = -\frac{2}{3}a$, we have that
$$P(-a) + P\left(\frac{8}{3}a\right) + P\left(-\frac{5}{3}a\right) = 2P\left(\frac{7}{3}a\right),$$
or
$$a_n \left[ 1 + \left(\frac{8}{3}\right)^{2n} + \left(\frac{5}{3}\right)^{2n} - 2\left(\frac{7}{3}\right)^{2n} \right] a^{2n} + \cdots + a_1 \left[ 1 + \left(\frac{8}{3}\right)^2 + \left(\frac{5}{3}\right)^2 - 2\left(\frac{7}{3}\right)^2 \right] a^2 = 0$$
for all $a \in \mathbb{R}$. Then all coefficients of the polynomial with variable $a$ are $0$.
If $n \ge 3$, it follows from $8^6 = 262\,144 > 235\,298 = 2 \times 7^6$ that $\left(\frac{8}{7}\right)^{2n} \ge \left(\frac{8}{7}\right)^6 > 2$. This implies
$$1 + \left(\frac{8}{3}\right)^{2n} + \left(\frac{5}{3}\right)^{2n} - 2\left(\frac{7}{3}\right)^{2n} > 0.$$
Hence $n \le 2$. Let $P(x) = a x^4 + \beta x^2$, with $a, \beta \in \mathbb{R}$.

We now show that $P(x) = a x^4 + \beta x^2$ satisfies the given equation.
Let $a, b, c$ be real numbers satisfying $ab + bc + ca = 0$. Then
$$\begin{align*} (a-b)^4 + (b-c)^4 + (c-a)^4 - 2(a+b+c)^4 &= \sum (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4) - 2(a^2 + b^2 + c^2)^2 \\ &= \sum (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4) - \\ & \qquad 2a^4 - 2b^4 - 2c^4 - 4a^2b^2 - 4a^2c^2 - 4b^2c^2 \\ &= \sum (-4a^3b + 2a^2b^2 - 4ab^3) \\ &= -4a^2(ab + ca) - 4b^2(bc + ab) - 4c^2(ca + bc) + \\ & \qquad 2(a^2b^2 + b^2c^2 + c^2a^2) \\ &= 4a^2bc + 4b^2ca + 4c^2ab + 2a^2b^2 + 2b^2c^2 + 2c^2a^2 \\ &= 2(ab + bc + ca)^2 = 0, \\ (a-b)^2 + (b-c)^2 + (c-a)^2 - 2(a+b+c)^2 \\ &= \sum (a^2 - 2ab + b^2) - 2\sum a^2 - 4\sum ab \\ &= 0. \end{align*}$$
Hence $P(x) = a x^4 + \beta x^2$ satisfies the given equation.