Find the cube of each of the following binomial expressions:$ \frac{3}{x}-\frac{2}{x^{2}} $

Given:

\( \frac{3}{x}-\frac{2}{x^{2}} \)

To do:

We have to find the cube of the given binomial expression.

Solution:

We know that,

$(a-b)^3=a^3 - b^3 - 3a^2b + 3ab^2$

Therefore,

$(\frac{3}{x}-\frac{2}{x^{2}})^{3}=(\frac{3}{x})^{3}-(\frac{2}{x^{2}})^{3}-3 \times(\frac{3}{x})^{2} \times(\frac{2}{x^{2}})+3 \times \frac{3}{x} \times(\frac{2}{x^{2}})^{2}$

$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-3 \times \frac{9}{x^{2}} \times \frac{2}{x^{2}}+3 \times \frac{3}{x} \times \frac{4}{x^{4}}$

$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\frac{54}{x^{4}}+\frac{36}{x^{5}}$

Hence, $(\frac{3}{x}-\frac{2}{x^{2}})^{3}=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\frac{54}{x^{4}}+\frac{36}{x^{5}}$.

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Updated on: 10-Oct-2022

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