Example 4 In Figure 1, there are 8 vertices, each with a real number. The real number at each vertex is exactly the average of the numbers at the 3 adjacent vertices (two vertices connected by a line segment are called adjacent vertices). Find
$$
a+b+c+d-(e+f+g+h)
$$
The following solution has appeared in a journal:
Given
$$
\begin{array}{l}
a=\frac{b+e+d}{3}, \\
b=\frac{a+f+c}{3}, \\
c=\frac{b+g+d}{3}, \\
d=\frac{c+h+a}{3}.
\end{array}
$$

Adding the four equations yields
$$
\begin{array}{l}
a+b+c+d \\
=\frac{1}{3}(2 a+2 b+2 c+2 d+e+f+g+h),
\end{array}
$$

which simplifies to $a+b+c+d-(e+f+g+h)=0$.
The above solution, though clever, has two drawbacks.
First, it is too coincidental. Adding the equations to get $a+b+c+d$ is expected, but the fact that it simplifies to $a+b+c+d-(e+f+g+h)=0$ is purely coincidental. If the problem were to find $2a+b+c+d-(2e+f+g+h)$, the above method would no longer work.

Second, the solution fails to reveal the essence of the problem, which is the relationship among the 8 numbers $a, b, c, d, e, f, g, h$.

To solve this problem, one should first consider a specific example. To make each vertex number equal to the average of the adjacent vertex numbers, the simplest example is to take
$$
a=b=c=d=e=f=g=h.
$$

In fact, this is the only possibility. Without loss of generality, assume $a$ is the largest of the 8 numbers. Since $a$ is the average of $b, e, d$, $b, e, d$ must also be the largest, i.e., equal to $a$. By the same reasoning, all 8 numbers must be equal. Thus, whether it is $a+b+c+d-(e+f+g+h)$ or $2a+b+c+d-(2e+f+g+h)$, the result is clearly 0.

Therefore, equations (1), (2), etc., are just superficial, while the equality of the 8 numbers is the true essence.