[Example 4.4.1] Given
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$

Find the value of \( x^{2}+y^{2}+z^{2}+w^{2} \).
We are accustomed to viewing the four equations in the problem as a system of four equations, i.e., a system of four equations with four unknowns \(x, y, z, w\). The conventional approach is to solve this system to find the values of \(x, y, z, w\), and then calculate \(x^{2}+y^{2}+z^{2}+w^{2}\). However, solving such a system of four quadratic equations is quite cumbersome. We notice that the problem ultimately requires the sum of the squares of \(x, y, z, w\), not the individual values of \(x, y, z, w\). Therefore, the final requirement is the overall value of \(x^{2}+y^{2}+z^{2}+w^{2}\). At this point, we should reconsider the four equations. If we do not treat \(x, y, z, w\) as unknowns, but instead treat \(2^{2}, 4^{2}, 6^{2}, 8^{2}\) as unknowns, the four equations become a single entity, i.e., a polynomial equation in \(t\) with roots \(2^{2}, 4^{2}, 6^{2}, 8^{2}\):

\[
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.
\]

This is a fractional equation in \(t\), which can be transformed into a quartic equation in \(t\):

\[
t^{4}-\left(x^{2}+y^{2}+z^{2}+w^{2}+84\right) t^{3}+a_{2} t^{2}+a_{3} t+a_{4}=0,
\]

Since \(2^{2}, 4^{2}, 6^{2}, 8^{2}\) are the roots of equation (2), we have:

\[
\left(t-2^{2}\right)\left(t-4^{2}\right)\left(t-6^{2}\right)\left(t-8^{2}\right)=0,
\]

which simplifies to:

\[
t^{4}-\left(2^{2}+4^{2}+6^{2}+8^{2}\right) t^{3}+a_{2} t^{2}+a_{3} t+a_{4}=0.
\]

By comparing the coefficients of \(t^{3}\) in equations (2) and (3), we get:

\[
\begin{array}{c}
x^{2}+y^{2}+z^{2}+w^{2}+84=2^{2}+4^{2}+6^{2}+8^{2}=120, \\
x^{2}+y^{2}+z^{2}+w^{2}=36.
\end{array}
\]

Thus, in the solution process, we did not treat the given four equations as equations in \(x, y, z, w\), but rather as a single quartic equation in \(2^{2}, 4^{2}, 6^{2}, 8^{2}\), thereby focusing on the overall value of \(x^{2}+y^{2}+z^{2}+w^{2}\).