Question
Expand the following:
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
✓
Answer
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}}\Big)^3 + \Big(\frac{\text{y}}{3}\Big)^3 + 3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
${\left[\text { Using identity, }(a-b)^3=a^3-b^3+3 a(-b)(a-b)\right]}$
$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x}=\left(\frac{1}{x}+\frac{y}{3}\right)$
$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x^2}+\frac{y^2}{3 x}$
${\left[\text { Using identity, }(a-b)^3=a^3-b^3+3 a(-b)(a-b)\right]}$
$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x}=\left(\frac{1}{x}+\frac{y}{3}\right)$
$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x^2}+\frac{y^2}{3 x}$
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