Task B-4.1. Determine the term in the expansion of the binomial

$$
\left(\sqrt[3]{\frac{2 x}{\sqrt{y}}}-\sqrt{\frac{y}{2 \sqrt[3]{x}}}\right)^{21}, \quad x, y>0
$$

which contains $x$ and $y$ with the same exponent.
## Solution.

In the expansion of

$$
\left(\sqrt[3]{\frac{2 x}{\sqrt{y}}}-\sqrt{\frac{y}{2 \sqrt[3]{x}}}\right)^{21}
$$

the general term is $\binom{21}{k} a^{21-k} b^{k}$, where

$$
\begin{aligned}
& a=\sqrt[3]{\frac{2 x}{\sqrt{y}}}=2^{\frac{1}{3}} x^{\frac{1}{3}} y^{-\frac{1}{6}} \\
& b=-\sqrt{\frac{y}{2 \sqrt[3]{x}}}=-2^{-\frac{1}{2}} x^{-\frac{1}{6}} y^{\frac{1}{2}}
\end{aligned}
$$

Now we have

$$
\begin{aligned}
\binom{21}{k} a^{21-k} b^{k} & =\binom{21}{k}\left(2^{\frac{1}{3}} x^{\frac{1}{3}} y^{-\frac{1}{6}}\right)^{21-k} \cdot\left(-2^{-\frac{1}{2}} x^{-\frac{1}{6}} y^{\frac{1}{2}}\right)^{k} \\
& =\binom{21}{k}(-1)^{k}\left(2^{\frac{21-k}{3}-\frac{k}{2}}\right)\left(x^{\frac{21-k}{3}-\frac{k}{6}}\right)\left(y^{\frac{k-21}{6}+\frac{k}{2}}\right)
\end{aligned}
$$

For $x$ and $y$ to have the same exponent, the following must hold

$$
\frac{21-k}{3}-\frac{k}{6}=\frac{k-21}{6}+\frac{k}{2}
$$

From this, we get $k=9$, so the desired term is

$$
-\binom{21}{9} \cdot 2^{-\frac{1}{2}} x^{\frac{5}{2}} y^{\frac{5}{2}}
$$