6.234. $\left\{\begin{array}{l}u^{-\frac{1}{2}} \cdot \sqrt[3]{u}+v^{-\frac{1}{2}} \cdot \sqrt[3]{v}=1.5 \\ u v=64\end{array}\right.$
## Solution.

Rewrite the system of equations as $\left\{\begin{array}{l}\frac{1}{\sqrt[6]{u}}+\frac{1}{\sqrt[6]{v}}=1.5 . \\ u v=64 .\end{array}\right.$

Let $\left\{\begin{array}{l}\sqrt[6]{u}=x>0, \\ \sqrt[6]{v}=y>0 .\end{array}\right.$ The system in terms of $x$ and $y$ becomes

$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=1.5, \\
(x y)^{6}=64,
\end{array} \text { from which we get for } x y \neq 0,\left\{\begin{array}{l}
x+y=1.5 x y, \\
x y= \pm 2 .
\end{array}\right.\right.
$$

This system is equivalent to two systems of equations:

$\left\{\begin{array}{l}x+y=1.5 x y, \\ x y=-2\end{array}\right.$ or $\left\{\begin{array}{l}x+y=1.5 x y, \\ x y=2 .\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=2 ;\end{array}\left\{\begin{array}{l}x_{2}=2, \\ y_{2}=1 .\end{array}\right.\right.\right.$

Then $\left\{\begin{array}{l}\sqrt[6]{u}=1, \\ \sqrt[6]{v}=2\end{array} \Rightarrow\left\{\begin{array}{l}u_{1}=1, \\ v_{1}=64 ;\end{array}\left\{\begin{array}{l}\sqrt[6]{u}=2, \\ \sqrt[6]{v}=1\end{array} \Rightarrow\left\{\begin{array}{l}u_{2}=64, \\ v_{2}=1 .\end{array}\right.\right.\right.\right.$

Answer: $(1 ; 64),(64 ; 1)$.