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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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