## Task B-2.3.

Solve the system of equations

$$
\begin{gathered}
x \sqrt{x} + y \sqrt{y} = 126 \\
x + y - 12(\sqrt{x} + \sqrt{y}) + 36 + 2 \sqrt{x y} = 0
\end{gathered}
$$
## Solution.

Let's write the given equations in a more convenient form.

Since $x \sqrt{x}+y \sqrt{y}=(\sqrt{x})^{3}+(\sqrt{y})^{3}$, the first equation becomes

$$
\begin{aligned}
& (\sqrt{x}+\sqrt{y}) \cdot(x-\sqrt{x y}+y)=126 \\
& (\sqrt{x}+\sqrt{y}) \cdot\left((\sqrt{x}+\sqrt{y})^{2}-3 \sqrt{x y}\right)=126
\end{aligned}
$$

The second equation becomes

$$
\begin{aligned}
& x+y-12(\sqrt{x}+\sqrt{y})+36+2 \sqrt{x y}=0 \\
& (x+2 \sqrt{x y}+y)-12(\sqrt{x}+\sqrt{y})+36=0 \\
& (\sqrt{x}+\sqrt{y})^{2}-12(\sqrt{x}+\sqrt{y})+36=0 \\
& (\sqrt{x}+\sqrt{y}-6)^{2}=0 \\
& \Rightarrow \sqrt{x}+\sqrt{y}=6
\end{aligned}
$$

Substituting the obtained equality into $(*)$ gives $6 \cdot(36-3 \cdot \sqrt{x y})=126$, or $\sqrt{x y}=5$. We solve the system

$$
\begin{gathered}
\sqrt{x}+\sqrt{y}=6 \\
\sqrt{x y}=5
\end{gathered}
$$

By introducing the substitution $\sqrt{x}=a, \sqrt{y}=b$, we get

$$
\begin{gathered}
a+b=6 \\
a b=5
\end{gathered}
$$

which gives $a=1, b=5$ or $a=5, b=1$, i.e., $x=1, y=25$ or $x=25, y=1$.