Let's determine

$$
\left(\frac{7 \sqrt[3]{x}}{2 \sqrt{y}}-\frac{\sqrt[3]{y}}{7 \sqrt[5]{x^{2}}}\right)^{8}
$$

the sixth term of the expansion.
Solution: The sixth term of the expansion according to the binomial theorem is:

$$
\begin{aligned}
& \binom{8}{5}\left(\frac{7 \sqrt[3]{x}}{2 \sqrt{y}}\right)^{3}\left(-\frac{\sqrt[3]{y}}{7 \sqrt[5]{x^{2}}}\right)^{5}=\binom{8}{3} \frac{7^{3} x}{2^{3} y \sqrt{y}}\left(-\frac{y \sqrt[3]{y^{2}}}{7^{5} x^{2}}\right)= \\
& =-\frac{8 \cdot 7 \cdot 6}{1 \cdot 2 \cdot 3} \cdot \frac{7^{3} x y \sqrt[3]{y^{2}}}{2^{3} 7^{5} x^{2} y \sqrt{y}}=-\frac{\sqrt[3]{y^{2}}}{7 x \sqrt{y}}=-\frac{\sqrt[6]{y^{4}}}{7 x \sqrt[6]{y^{3}}}=-\frac{\sqrt[6]{y}}{7 x}
\end{aligned}
$$

Károly Horváth (Vác, Sztáron S. g. III. o. t.)