Find the Cube of the Following Binomial Expression
\[ \frac{1}{x}+\frac{y}{3} \]
Solution:
Using identity:
\[ (a+b)^3 = a^3+b^3+3ab(a+b) \]
\[ \left(\frac{1}{x}+\frac{y}{3}\right)^3 \]
\[ = \left(\frac{1}{x}\right)^3 + \left(\frac{y}{3}\right)^3 + 3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right) \left(\frac{1}{x}+\frac{y}{3}\right) \]
\[ = \frac{1}{x^3} + \frac{y^3}{27} + \frac{y}{x} \left(\frac{1}{x}+\frac{y}{3}\right) \]
\[ = \frac{1}{x^3} + \frac{y^3}{27} + \frac{y}{x^2} + \frac{y^2}{3x} \]