6.1.9 Find all real-coefficient polynomials $p(x)$ such that for any real numbers $a, b, c$ satisfying $a b + b c + c a = 0$, we have
$$
p(a-b) + p(b-c) + p(c-a) = 2 p(a+b+c).
$$
Let $a=b=0$, then for any $c \in \mathbf{R}$, we have $a b+b c+c a=0$, thus
$$
p(0)+p(-c)+p(c)=2 p(c),
$$

which implies $p(c)=p(-c)+p(0)$. Setting $c=0$ in this equation, we get $p(0)=0$, so for any $c \in \mathbf{R}$, we have $p(c)=p(-c)$. Therefore, $p(x)$ is an even function, and we can assume $p(x)=a_{n} x^{2 n}+a_{n-1} x^{2(n-1)}+\cdots +a_{1} x^{2}$ (here we use $p(0)=0$), where $a_{1}, \cdots, a_{n} \in \mathbf{R}$.

Now, in (1), let $(a, b, c)=(6 x, 3 x, -2 x)$, then $a b+b c+c a=0$, thus $p(3 x)+ p(5 x)+p(-8 x)=2 p(7 x)$ (2). Combining $p(x)=\sum_{i=1}^{n} a_{i} x^{2 i}$, comparing the coefficients of the corresponding terms of $x$ on both sides of (2), we get
$$
\left(3^{i}+5^{i}+(-8)^{i}-2 \cdot 7^{i}\right) a_{i}=0, \quad i=1,2, \cdots, n.
$$

When $i$ is odd, $3^{i}+5^{i}+(-8)^{i}-2 \cdot 7^{i}=3^{i}+5^{i}-8^{i}-2 \cdot 7^{i} \geqslant 8^{6}-2 \cdot 7^{6}$, and $\left(\frac{8}{7}\right)^{6}=\left(1+\frac{1}{7}\right)^{6}>1+\mathrm{C}_{6}^{1} \cdot \frac{1}{7}+\mathrm{C}_{6}^{2} \cdot \frac{1}{7^{2}}=1+\frac{6}{7}+\frac{15}{49}>2$. Therefore, at this time, $3^{i}+5^{i}+(-8)^{i}-2 \cdot 7^{i}>0$, so $a_{i}=0$. This indicates: $p(x)=a_{2} x^{2}+a_{1} x^{4}$, where $a_{2}, a_{4} \in \mathbf{R}$.
On the other hand, for any $\alpha, \beta \in \mathbf{R}$, if $p(x)=\alpha x^{2}+\beta x^{4}$, then
$$
p(a-b)+p(b-c)+p(c-a)=\alpha \sum(a-b)^{2}+\beta \sum(a-b)^{4},
$$

for any real numbers $a, b, c$ satisfying $a b+b c+c a=0$, let $x=a-b, y=b-c, z=c-a$, then $x+y+z=0$, and
$$
x^{2}+y^{2}+z^{2}=2(a+b+c)^{2}-6(a b+b c+c a)=2(a+b+c)^{2}.
$$

From $x+y+z=0$ and $x^{2}+y^{2}+z^{2}=2(a+b+c)^{2}$, we know $x y+y z+z x=-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)=-(a+b+c)^{2}$, and
$$
\begin{array}{l}
(x y)^{2}+(y z)^{2}+(z x)^{2}=(x y+y z+z x)^{2}-2 x y z(x+y+z)=(a+b+c)^{4}, \\
x^{4}+y^{4}+z^{4}=\left(x^{2}+y^{2}+z^{2}\right)^{2}-2\left((x y)^{2}+(y z)^{2}+(z x)^{2}\right)=2(a+b+c)^{4},
\end{array}
$$

Therefore, combining (3) and $p(x)=\alpha x^{2}+\beta x^{4}$, we know (1) holds.
In summary, $p(x)=\alpha x^{2}+\beta x^{4}, \alpha, \beta \in \mathbf{R}$.