Question
Find the cube the following binomial expressions: $4-\frac{1}{3\text{x}}$
✓
Answer
Given, $4-\frac{1}{3\text{x}}$ The above equation is in the form of $(a - b)^3= a^3 - b^3 - 3ab(a - b)$
We know that, $\text{a} = 4, \text{b} =\frac{1}{3\text{x}}$ By using $(a - b)^3$
formula $\Big(4-\frac{1}{3\text{x}}\Big)^3$
$=4^3-\Big(\frac{1}{3\text{x}}\Big)^3-3(4)\Big(\frac{1}{3\text{x}}\Big)\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{12}{3\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{4}{\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\Big(\frac{4}{3\text{x}}\times4\Big)+\Big(\frac{4}{3\text{x}}\times\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
Hence The cube of $\Big(4-\frac{1}{3\text{x}}\Big)^3=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
We know that, $\text{a} = 4, \text{b} =\frac{1}{3\text{x}}$ By using $(a - b)^3$
formula $\Big(4-\frac{1}{3\text{x}}\Big)^3$
$=4^3-\Big(\frac{1}{3\text{x}}\Big)^3-3(4)\Big(\frac{1}{3\text{x}}\Big)\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{12}{3\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{4}{\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\Big(\frac{4}{3\text{x}}\times4\Big)+\Big(\frac{4}{3\text{x}}\times\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
Hence The cube of $\Big(4-\frac{1}{3\text{x}}\Big)^3=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
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