Expand the expression $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$

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