The value of $x^2+y^2+z^2+w^2$
Solution 1:
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*}
Solution 2:
Set up a matrix equation $$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{bmatrix} = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$$ and then solve. You can essentially ignore the fact that your variables are squared, and just assume that means they must all be nonnegative.
Solution 3:
I multiplied the equations out to get:
$$\begin{pmatrix} -4725 & 2835 & 675 & 315 \\ 2079 & 4455 & -3465 & -945 \\ -3861 & -5005 & -12285 & 10395 \\ 32175 & 36855 & 51975 & 135135 \\ \end{pmatrix} \begin{pmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{pmatrix} = \begin{pmatrix} -14175 \\ 31185 \\ -135135 \\ 2027025 \\ \end{pmatrix}$$
Then inverted the matrix and solved to get:
$$\begin{pmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{pmatrix} = \begin{pmatrix} 10.76660156 \\ 10.15136719 \\ 8.797851563 \\ 6.284179688 \end{pmatrix} $$
giving $$x^2+y^2+z^2+w^2=36$$ - so there probably is an algebraic short cut to that.