Question
Find the cube of the following binomial expressions: $\frac{1}{\text{x}}+\frac{\text{y}}{3}$
✓
Answer
Given, $\frac{1}{\text{x}}+\frac{\text{y}}{3}$ The above equation is in the form of $(a + b)^3= a^3 + b^3 + 3ab(a + b)$
We know that,$ \text{a} =\frac{1}{\text{x}},\text{b} =\frac{\text{y}}{3}$ By using $(a + b)^3$ ^formula $\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
$=\Big(\frac{1}{\text{x}}\Big)^3+\Big(\frac{\text{y}}{3}\Big)^3+3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+3\times\frac{1}{\text{x}}\times\frac{\text{y}}{3}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\Big(\frac{\text{y}}{\text{x}}\times\frac{1}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\times\text{y}^3\Big)$
$=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
Hence $\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
We know that,$ \text{a} =\frac{1}{\text{x}},\text{b} =\frac{\text{y}}{3}$ By using $(a + b)^3$ ^formula $\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
$=\Big(\frac{1}{\text{x}}\Big)^3+\Big(\frac{\text{y}}{3}\Big)^3+3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+3\times\frac{1}{\text{x}}\times\frac{\text{y}}{3}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\Big(\frac{\text{y}}{\text{x}}\times\frac{1}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\times\text{y}^3\Big)$
$=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
Hence $\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
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