$\begin{array}{l}\text { 15. Find the value of } x^{2}+y^{2}+z^{2}+w^{3} \text { . If } \\ \frac{x^{2}}{\varepsilon^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}} \\ +\frac{w^{2}}{2^{2}-7^{2}}=1, \\ \frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}} \\ +\frac{w^{2}}{4^{2}-7^{2}}=1, \\ \frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\ \frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}} \\ +\frac{w^{2}}{8^{2}-7^{2}}=1 .\end{array}$
Solving for $\mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{w}$ to satisfy the given system of equations is equivalent to $t=4,16,36,64$ satisfying 
$$
\frac{x^{2}}{t-1}+\frac{y^{2}}{t-9}+\frac{z^{2}}{t-25}=\frac{w^{2}}{t-49}=1. \quad (1)
$$

Multiplying both sides of (1) by the denominators, we find that for meaningful $t$ (i.e., $t \neq 1,9,25,49$), (1) is equivalent to the polynomial equation:
$$
\begin{array}{c}
(t-1)(t-9)(t-25)(t-49) \\
-x^{2}(t-9)(t-25)(t-49)-y^{2}(t-1) \\
-(t-25)(t-49)-z^{2}(t-1)(t-9)(t-
\end{array}
$$
49) $-w^{2}(t-1)(t-9)(t-25)=0. \quad (2)$

The left side is a fourth-degree polynomial in $\mathrm{t}$. Since $\mathrm{t}=4,16,36,64$ are four known roots of this equation, and a fourth-degree polynomial can have at most four roots, these are all the roots of equation (2). Therefore, (2) is equivalent to $(t-4)(t-16)(t-36)(t-64)=0. \quad (3)$

In (2) and (3), the coefficient of $t^{4}$ is 1, so the coefficients of the other powers of $t$ should also be equal. In particular, the coefficients of $t^{3}$ in both equations should be equal. By the relationship between roots and coefficients and comparing terms, we get
$$
\begin{array}{l}
1+9+25+49+x^{2}+y^{2}+z^{2}+w^{2} \\
=4+16+36+64.
\end{array}
$$

From this, we find
$$
\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+\mathrm{w}^{2}=36.
$$
(Supplied by Ma Xiwen and Zhou Chunlu, Beijing Mathematical Society)