Question

solve this problem in explanation
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Answer 1

➤ Answer :-

${\to \tt \dfrac{{\bigg(\dfrac{( - 3)}{5}}^{3} \bigg) \times{ \bigg(\dfrac{9}{25} }^{2} \bigg) \times {\bigg( \dfrac{( - 18)}{125}}^{0} \bigg)}{\dfrac{( - 27)}{125} \times \dfrac{( - 3)}{5}}}$

Distribute the powers to both the parts of fraction in the numerators.

${\to \tt \dfrac{\bigg(\dfrac{( - 3) \times ( - 3) \times ( - 3)}{5 \times 5 \times 5} \bigg) \times \bigg(\dfrac{9 \times 9}{25 \times 25} \bigg) \times 1}{\dfrac{( - 27) \times ( - 3)}{125 \times 5}}}$

Simplify each of the given values by multiplying them.

${\to \tt \dfrac{\bigg(\dfrac{( - 27)}{125} \bigg) \times \bigg(\dfrac{81}{625} \bigg)}{ \dfrac{81}{625}}}$

Simplify each of the numerators and denominators by cancellation method.

${\to \tt \dfrac{\dfrac{( - 27)}{125} \times \cancel \dfrac{81}{625}}{\cancel \dfrac{81}{625}} = \dfrac{\dfrac{( - 27)}{125} \times 1}{1}}$

${\to \tt \dfrac{( - 27)}{125} = \boxed{\tt \bigg({\dfrac{( - 3)}{5}} \bigg)^{3}}}$

$\Huge\therefore$ The answer of the question is ${\tt \bigg({\dfrac{( - 3)}{5}} \bigg)^{3}}$.

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$\dashrightarrow$ Some related equations :-

${\to \sf {a}^{m} \times {a}^{n} = {a}^{m + n}}$

${\to \sf {a}^{m} \div {a}^{n} = {a}^{m - n}}$

${\to \sf \bigg( {a}^{m} \bigg)^{n} = {a}^{m \times n}}$

${\to \sf \dfrac{ {a}^{m}}{ {b}^{m}} = \bigg( {\dfrac{a}{b}}\bigg)^{m}}$

${\to \sf {a}^{0} = 1}$

${\to \sf {a}^{ - 1} = \dfrac{1}{a}}$