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

Find the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
The analysis and solution: $x, y, z, w$ can satisfy the given system of equations if and only if $t=4,16,36,64$ satisfy the equation
$$
\frac{x^{2}}{t-1}+\frac{y^{2}}{t-9}+\frac{z^{2}}{t-25}+\frac{w^{2}}{t-49}=1
$$

Clearing the denominators, when $t \neq 1,9,25,49$, the equation (1) is equivalent to the equation
$$
\begin{array}{l}
(t-1)(t-9)(t-25)(t-49)-x^{2}(t-9)(t-25)(t-49)-y^{2}(t-1)(t-25)(t-49)- \\
z^{2}(t-1)(t-9)(t-49)-w^{2}(t-1)(t-9)(t-25)=0
\end{array}
$$

Equation (2) is a quartic equation in $t$, and $t=4,16,36,64$ are the 4 known roots of this equation, which are all its roots. Therefore, equation (2) is equivalent to $(t-4)(t-16)(t-36)(t-64)=0$.
Since the coefficients of $t^{4}$ in equations (2) and (3) are both 1, the coefficients of the other powers of $t$ should also be equal. Comparing the coefficients of $t^{3}$, we get
$$
1+9+25+49+x^{2}+y^{2}+z^{2}+w^{2}=4+16+36+64 .
$$

Thus, we obtain
$$
x^{2}+y^{2}+z^{2}+w^{2}=36 .
$$