Question
Find the cube of the following binomial expressions:
$\frac{3}{\text{x}}-\frac{2}{\text{x}^2}$
$\frac{3}{\text{x}}-\frac{2}{\text{x}^2}$
✓
Answer
Given,$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$
The above equation is in the form of $(a - b)^3= a^3 - b^3 - 3ab(a - b)$
We know that, $\text{a}=\frac{3}{\text{x}},\text{b}=\frac{2}{\text{x}^2}$
By using $(a - b)^3$ formula
$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$
$=(3\text{x})^3-\Big(\frac{2}{\text{x}^2}\Big)^3-3\Big(\frac{3}{\text{x}}\Big)\Big(\frac{2}{\text{x}^2}\Big)\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-3\times\frac{3}{\text{x}}\times\frac{2}{\text{x}^2}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{18}{\text{x}^3}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\Big(\frac{18}{\text{x}^3}\times\frac{3}{\text{x}}\Big)+\Big(\frac{18}{\text{x}^3}\times\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
Hence $\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
The above equation is in the form of $(a - b)^3= a^3 - b^3 - 3ab(a - b)$
We know that, $\text{a}=\frac{3}{\text{x}},\text{b}=\frac{2}{\text{x}^2}$
By using $(a - b)^3$ formula
$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$
$=(3\text{x})^3-\Big(\frac{2}{\text{x}^2}\Big)^3-3\Big(\frac{3}{\text{x}}\Big)\Big(\frac{2}{\text{x}^2}\Big)\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-3\times\frac{3}{\text{x}}\times\frac{2}{\text{x}^2}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{18}{\text{x}^3}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\Big(\frac{18}{\text{x}^3}\times\frac{3}{\text{x}}\Big)+\Big(\frac{18}{\text{x}^3}\times\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
Hence $\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
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