Question

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

Answer 1

By using Binomial Theorem, the expression $(\frac{x}{3} + \frac{1}{x})^5$ can be expanded as

$(\frac{x}{3} + \frac{1}{x})^5$ = $^5C_0(\frac{x}{3})^5 + ^5C_1(\frac{x}{3})^4(\frac{1}{x}) + ^5C_2 (\frac{x}{3})^3(\frac{1}{x})^2 + ^5C_3(\frac{x}{3})^2(\frac{1}{x})^3 + ^5C_4(\frac{x}{3})(\frac{1}{x})^4+ ^5C_5(\frac{1}{x})^5$

= $\frac{x^5}{ 243} + 5(\frac{x^4}{81})(\frac{1}{x}) + 10 (\frac{x^3}{27})(\frac{1}{x^2}) + 10 (\frac{x^2}{9})(\frac{1}{x^3}) + 5 (\frac{x}{3})(\frac{1}{x^4}) +(\frac{1}{x^5})$

= $\frac{x^5}{ 243} + \frac{5x^3}{81} + \frac{10x}{ 27} + \frac{10}{9x} + \frac{5}{ 3x^3} + \frac{1}{x^5}$.