Question

value of $\sqrt[5]{32^{-3} }$
fast please

Answer 1

Answer:

$given \: that = \sqrt[5]{ {32}^{ - 3} } \\ \: = > \sqrt[5]{ {(2}^{5} } ) ^{ - 3} \\ \: = > \sqrt[5]{( {2}^{5 \times - 3} } ) \\ \: = > \sqrt[5]{ {2}^{ - 15} } \\ \: = > ( {2}^{ - 15} )^{ \frac{1}{5} } \\ \: (since \: \sqrt[n]{a} = ( {a}^{ \frac{1}{n} } ) \\ \: = > {2}^{ \frac{ - 15}{5} } \\ \: (since \: {a}^{m} \div {a}^{n} = {a}^{m - n)} \\ \: = > {2 }^{ - 3} \\ \: = > \frac{1}{ {2}^{3} } \\ \: (since \: {a}^{ - n} = \frac{1}{ {a}^{n} } ) \\ \: = > \frac{1}{8} \\ \: answer \: is \: \frac{1}{8}$

Answer 2

Answer:

1/8 is the answer for your question