The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\frac{w^2}{6^2 - 7^2} =1$$
$$\frac{x^2}{8^2 - 1^2} +\frac{y^2}{8^2 - 3^2} +\frac{z^2}{8^2 - 5^2} +\frac{w^2}{8^2 - 7^2} =1$$
My work
$$\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1$$
where $t $ satisfy $4,16,36,64$
$$f(t)=0$$
$$f(t) = (t – 1)(t – 9)(t – 25)(t – 49)–x^2(t – 9)(t – 25)(t – 49) –y^2(t – 1)(t – 25)(t – 49) – z^2(t–1)(t–9)(t–49) – w^2(t–1)(t–9)(t–25)$$
then I compared the coefficient with different value of $t$ .
I want to know that is there any easier alternative methods for this .

Expanding 
\begin{equation*}
\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 
\end{equation*}
we get 
\begin{equation*}
(t-1^2)(t-3^2)(t-5^2)(t-7^2) - (t-3^2)(t-5^2)(t-7^2)x^2 - \text{ similar terms } = 0 
\end{equation*}
This biquadratic in $t$ has four roots $2^2, 4^2, 6^2, 8^2$ and coefficient of $t^3$ is $-(2^2+4^2+6^2+8^2)$. The coefficient of $t^3$ is also given by
\begin{equation*}
-(1^2+3^2+5^2+7^2) - (x^2+y^2+z^2+w^2)
\end{equation*}
and hence
\begin{equation*}
-(1^2+3^2+5^2+7^2) - (x^2+y^2+z^2+w^2) = -(2^2+4^2+6^2+8^2)
\end{equation*}
Hence we have
\begin{equation*}
x^2+y^2+z^2+w^2 = (2^2+4^2+6^2+8^2)-(1^2+3^2+5^2+7^2)=36
\end{equation*}
