Example 8. (I $150-17-3$ )
Find all polynomials in two variables that satisfy the following conditions:
1) $P$ is a homogeneous polynomial of degree $n$, i.e., for all real numbers $t$, $x$, $y$, we have
$$
P(t x, t y)=t^{\mathbb{n}} P(x, y) \text {; }
$$
2) For all real numbers $a, b, c$ we have
$$
\begin{array}{l}
P(a+b, c)+P(b+c, a)+P(c+a, b)=0 \text {; } \\
\quad ; \text {; } P(1,0)=1 .
\end{array}
$$
Solving H constant multiple must not satisfy 2) and 3) simultaneously, hence $n \geqslant 1$.

In 2), let $a=b, c=-2a$, then by the homogeneity of $P$ we have
$$
\begin{aligned}
0 & =P(2a, -2a) + P(-a, a) + P(-a, a) \\
& =\left[(-2)^{n} + 2\right] P(-a, a).
\end{aligned}
$$

If $n>1$, then $P(-a, a)=0$, implying that $P(x, y)$ can be divided by $x+y$. Set
$$
P(x, y) = (x+y) P_{1}(x, y),
$$

then from 2) we get
$$
\begin{array}{l}
P_{1}(a+b, c)(a+b+c) + P_{1}(b+c, a) \\
\cdot (a+b+c) + P_{1}(c+a, b)(a+b+c) \\
= 0,
\end{array}
$$

which simplifies to
$$
\begin{array}{l}
P_{1}(a+b, c) + P_{1}(b+c, a) + P_{1}(c+a, b) \\
\quad = 0.
\end{array}
$$
$P_{1}(x, y)$ is still a homogeneous polynomial of degree $n-1$, and
$$
P_{1}(1,0) = P_{1}(1,0)(1+0) = P(1,0) = 1,
$$

meaning $P_{1}(x, y)$ still satisfies conditions 1) - 3). If $n-1>1$, similarly, we can construct another homogeneous polynomial of degree $n-2$, $P_{2}(x, y)$, which satisfies conditions 1) - 3), and
$$
P_{1}(x, y) = (x+y) P_{2}(x, y).
$$

Continuing this process, we will eventually obtain a linear polynomial
$$
R_{n-1}(x, y) = A x + B y.
$$
From condition 2), let $a=1, b=c=0$, we get
$$
2 + B = 0, \quad B = -2.
$$

Thus,
$$
P(x, y) = (x+y)^{n-1}(x-2y).
$$

It is not difficult to verify that this polynomial meets our requirements. From the above reasoning, we can conclude that all polynomials satisfying conditions 1) - 3) are of this form.