Expand the expression $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$
By using Binomial Theorem, the expression $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$ can be expanded as
$\left(\frac{x}{3}+\frac{1}{x}\right)^{5}={ }^{5} C_{0}\left(\frac{x}{3}\right)^{5}+{ }^{5} C_{1}\left(\frac{x}{3}\right)^{4}\left(\frac{1}{x}\right)+{ }^{5} C_{2}\left(\frac{x}{3}\right)^{3}\left(\frac{1}{x}\right)^{2}$
$+{ }^{5} \mathrm{C}_{3}\left(\frac{\mathrm{x}}{3}\right)^{2}\left(\frac{1}{\mathrm{x}}\right)^{3}+{ }^{5} \mathrm{C}_{4}\left(\frac{\mathrm{x}}{3}\right)\left(\frac{1}{\mathrm{x}}\right)^{4}+{ }^{5} \mathrm{C}_{5}\left(\frac{1}{\mathrm{x}}\right)^{5}$
$=\frac{x^{5}}{243}+5\left(\frac{x^{4}}{81}\right)\left(\frac{1}{x}\right)+10\left(\frac{x^{3}}{27}\right)\left(\frac{1}{x^{2}}\right)+10\left(\frac{x^{2}}{9}\right)\left(\frac{1}{x^{3}}\right)+5\left(\frac{x}{3}\right)\left(\frac{1}{x^{4}}\right)+\frac{1}{x^{3}}$
$=\frac{x^{5}}{243}+\frac{5 x^{3}}{81}+\frac{10 x}{27}+\frac{10}{9 x}+\frac{5}{3 x^{3}}+\frac{1}{x^{5}}$