6.190. $\left\{\begin{array}{l}x y=a, \\ y z=b, \quad a b c>0 . \\ z x=c,\end{array}\right.$
Solution.

From the third equation, we find $z=\frac{c}{x}$ and substitute it into the second equation $\left\{\begin{array}{l}x y=a \\ y \cdot \frac{c}{x}=b, \\ z=\frac{c}{x} .\end{array}\right.$

From the second equation, we find $y=\frac{b}{c} x$ and substitute it into the first equation:

$$
\left\{\begin{array} { l } 
{ x \cdot \frac { b } { c } x = a , } \\
{ y = \frac { b } { c } x } \\
{ z = \frac { c } { x } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 2 } = \frac { a c } { b } , } \\
{ y = \frac { b } { c } x } \\
{ z = \frac { c } { x } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x= \pm \sqrt{\frac{a c}{b}} \\
y=\frac{b}{c} x \\
z=\frac{c}{x}
\end{array}\right.\right.\right.
$$

The last system is equivalent to the combination of two systems:

1) $\left\{\begin{array}{l}x=\sqrt{\frac{a c}{b}}, \\ y=\frac{b}{c} x, \\ z=\frac{c}{x},\end{array}\left\{\begin{array}{l}x=\sqrt{\frac{a c}{b}}, \\ y=\sqrt{\frac{a b}{c}}, \text { or } \\ z=\sqrt{\frac{b c}{a}},\end{array}\right.\right.$
2) $\left\{\begin{array}{l}x=-\sqrt{\frac{a c}{b}}, \\ y=\frac{b}{c} x, \\ z=\frac{c}{x},\end{array} \quad\right.$ from which $\left\{\begin{array}{l}x=-\sqrt{\frac{a c}{b}}, \\ y=-\sqrt{\frac{a b}{c}}, \\ z=-\sqrt{\frac{b c}{a}} .\end{array}\right.$

Answer: $x_{1,2}= \pm \sqrt{\frac{a c}{b}} ; y_{1,2}= \pm \sqrt{\frac{a b}{c}} ; z_{1,2}= \pm \sqrt{\frac{b c}{a}}$.