4・ 130 Solve the system of equations
$$\left\{\begin{array}{l}
x^{2}=a+(y-z)^{2} \\
y^{2}=b+(z-x)^{2} \\
z^{2}=c+(x-y)^{2}
\end{array}\right.$$
[Solution] Move all variables in the system of equations to the left and factorize, we get
$$\left\{\begin{array}{l}
(x-y+z)(x+y-z)=a, \\
(x+y-z)(-x+y+z)=b, \\
(-x+y+z)(x-y+z)=c .
\end{array}\right.$$

If \( a b c > 0 \), multiplying the above three equations and taking the square root, we get
$$(x-y+z)(x+y-z)(-x+y+z)= \pm \sqrt{a b c}$$

Dividing (1) by each of the above equations, the two sets of solutions for the system of equations are:
$$\begin{array}{l}
x= \pm \frac{b+c}{2 b c} \sqrt{a b c} \\
y= \pm \frac{a+c}{2 a c} \sqrt{a b c} \\
z= \pm \frac{a+b}{2 a b} \sqrt{a b c}
\end{array}$$

The signs in the fractions are taken as all positive or all negative simultaneously.
If \( a b c < 0 \) or \( a b c = 0 \), similar discussions can be made.