Solve the following system of equations:

$$
\begin{aligned}
& x^{2}+y \sqrt{x y}=336 \\
& y^{2}+x \sqrt{x y}=112
\end{aligned}
$$
By factoring out $\sqrt{x}$ from the left side of the first equation and $\sqrt{y}$ from the left side of the second equation:

$$
\begin{aligned}
& \sqrt{x}(x \sqrt{x}+y \sqrt{y})=3 \cdot 112 \\
& \sqrt{y}(y \sqrt{y}+x \sqrt{x})=112
\end{aligned}
$$

If we divide the first equation by the second:

$$
\frac{\sqrt{x}}{\sqrt{y}}=3, \quad \text { or } \quad x=9 y
$$

Substituting this value of $x$ into the second equation:

$$
\begin{aligned}
& y^{2}+27 y^{2}=4 \cdot 28 \\
& y^{2}=4 \\
& y= \pm 2 \quad \text { and } \quad x= \pm 18
\end{aligned}
$$

but only the values $x=18$, and $y=2$ satisfy the system of equations.

Solved by: Barabás Gy., Dancs I., Kántor S., Müller Z., Révész P., Zatykó L., Zobor E.