Example 9 Find all real polynomials $P(x)$ such that for all real numbers $a, b, c$ satisfying $ab + bc + ca = 0$, we have $P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c)$.
(IMO - 45 Problem)
Given that for any $a, b, c \in \mathbf{R}$, satisfying $a b + b c + c a = 0$, we have
$$
P(a - b) + P(b - c) + P(c - a) = 2 P(a + b + c).
$$

By setting $a = b = c = 0$ in (1), we get $P(0) = 0$;
By setting $b = c = 0$ in (1), we get $P(-a) = P(a)$ for any real number $a$.
Therefore, all the coefficients of the odd powers of $P(x)$ are 0 (i.e., $P(x)$ is an even function).
Assume $P(x) = a_{n} x^{2 n} + \cdots + a_{1} x^{2}, a_{n} \neq 0$.
By setting $b = 2a$ and $c = -\frac{2}{3}a$ in (1), we have $P(-a) + P\left(\frac{8}{3}a\right) + P\left(-\frac{5}{3}a\right) = 2 P\left(\frac{7a}{3}\right)$. That is, 
$$
a_{n}\left[1 + \left(\frac{8}{3}\right)^{2 n} + \left(\frac{5}{3}\right)^{2 n} - 2\left(\frac{7}{3}\right)^{2 n}\right] \cdot a^{2 n} + \cdots + a_{1}\left[1 + \left(\frac{8}{3}\right)^{2} + \left(\frac{5}{3}\right)^{2} - 2\left(\frac{7}{3}\right)\right] a^{2} = 0,
$$
for all $a \in \mathbf{R}$, hence all the coefficients of the polynomial in $a$ are 0.
When $n \geqslant 3$, by $8^{6} = 262144 > 235298 = 2 \cdot 7^{6}$, we know $\left(\frac{8}{7}\right)^{2 n} \geqslant \left(\frac{8}{7}\right)^{6} > 2$.
Thus, $1 + \left(\frac{8}{3}\right)^{2 n} + \left(\frac{5}{3}\right)^{2 n} - 2\left(\frac{7}{3}\right)^{2 n} > 0$.
Therefore, $n \leqslant 2$, so we can set $P(x) = \alpha x^{4} + \beta x^{2}, \alpha, \beta \in \mathbf{R}$.
Now, we verify that $P(x) = \alpha x^{4} + \beta x^{2}$ satisfies the requirement.
Let $a, b, c \in \mathbf{R}$, and $a b + b c + c a = 0$, then
$$
\begin{array}{l}
(a - b)^{4} + (b - c)^{4} + (c - a)^{4} - 2(a + b + c)^{4} \\
= \sum\left(a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + b^{4}\right) - 2\left(a^{2} + b^{2} + c^{2}\right)^{2} \\
= \sum\left(a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + b^{4}\right) - 2 a^{4} - 2 b^{4} - 2 c^{4} - 4\left(a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}\right) \\
= -4 a^{2}(a b + c a) - 4 b^{2}(b c + a b) - 4 c^{2}(c a + b c) + 2\left(a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}\right) \\
= 4 a^{2} b c + 4 b^{2} c a + 4 c^{2} a b + 2\left(a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}\right) = 2(a b + b c + c a)^{2} = 0.
\end{array}
$$

Thus, $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} - 2(a + b + c)^{2}$
$$
= \sum\left(a^{2} - 2 a b + b^{2}\right) - 2 \sum a^{2} - 4 \sum a b = 0.
$$

Therefore, $P(x) = \alpha x^{4} + \beta x^{2}$ satisfies the requirement.