Test $\mathrm{L}$ Let real numbers $x, y, z, w$ satisfy
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}\right.
$$

Find $x^{2}+y^{2}+z^{2}+w^{2}$.
(American Mathematical Invitational)
We first transform this problem into a general $n$-variable linear equation system similar in structure:
$$
\left\{\begin{array}{c}
\frac{x_{1}}{\lambda_{1}-a_{1}}+\frac{x_{2}}{\lambda_{1}-a_{2}}+\cdots+\frac{x_{n}}{\lambda_{1}-a_{n}}=1, \\
\frac{x_{1}}{\lambda_{2}-a_{1}}+\frac{x_{2}}{\lambda_{2}-a_{2}}+\cdots+\frac{x_{n}}{\lambda_{2}-a_{n}}=1, \\
\vdots \\
\frac{x_{1}}{\lambda_{n}-a_{1}}+\frac{x_{2}}{\lambda_{n}-a_{2}}+\cdots+\frac{x_{n}}{\lambda_{n}-a_{n}}=1,
\end{array}\right.
$$

where $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}$ and $a_{1}, a_{2}, \cdots, a_{n}$ are all distinct real numbers. Then we find the sum $\sum_{i=1}^{n} x_{i}$.
Let $\quad f(x)=\prod_{i=1}^{n}\left(x-a_{i}\right)-\prod_{i=1}^{n}\left(x-\lambda_{i}\right)$,

then
$$
f\left(\lambda_{k}\right)=\prod_{i=1}\left(\lambda_{k}-a_{i}\right) \quad(k=1,2, \cdots, n) .
$$

In the Lagrange polynomial, let $x=\lambda_{k}$, and divide both sides by $f\left(\lambda_{k}\right)$, we get
$$
\sum_{i=1}^{n} \frac{A_{i}}{\lambda_{k}-a_{i}}=1,
$$

where $A_{j}=f\left(a_{j}\right) / \prod_{\substack{1 \leq i \leq n \\ i \neq j}}\left(a_{j}-a_{i}\right) \quad(j, k=1,2, \cdots, n)$.
By uniqueness, the original system of equations has a unique solution $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$. Note that $f(x)$ can be written as
$$
\begin{aligned}
f(x) & =\prod_{i=1}^{n}\left(x-a_{i}\right)-\prod_{i=1}^{n}\left(x-\lambda_{i}\right) \\
& =\left[\sum_{i=1}^{n}\left(\lambda_{i}-a_{i}\right)\right] x^{n-1} .
\end{aligned}
$$

By the polynomial identity theorem, we have
$$
\sum_{j=1}^{n} f\left(a_{j}\right) / \prod_{\substack{1 \leq i \leq n \\ i \neq j}}\left(a_{j}-a_{i}\right)=\sum_{i=1}^{n}\left(\lambda_{i}-a_{i}\right) .
$$

That is,
$$
\sum_{i=1}^{n} A_{i}=\sum_{i=1}^{n} \lambda_{i}-\sum_{i=1}^{n} a_{i} .
$$

Returning to the original problem, we get
$$
\begin{aligned}
x^{2}+y^{2}+z^{2}+w^{2}= & \left(2^{2}+4^{2}+6^{2}+8^{2}\right) \\
& -\left(1^{2}+3^{2}+5^{2}+7^{2}\right)=36 .
\end{aligned}
$$

This solution method is quite general.