What is ((20x^5y^2)/(5x^{-3)y^7})^{-3} ?

The solution to ((20x^5y^2)/(5x^{-3)y^7})^{-3} is (y^{15})/(64x^{24)}

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simplify ((20x^5y^2)/(5x^{-3)y^7})^{-3}

simplify

(\frac{20 x ^{5} y ^{2}}{5 x ^{- 3} y ^{7}})^{- 3}

\frac{y ^{15}}{64 x ^{24}}

Solution steps

(\frac{20 x ^{5} y ^{2}}{5 x ^{- 3} y ^{7}})^{- 3}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

(\frac{20 x ^{5} y ^{2}}{5 x ^{- 3} y ^{7}})^{- 3} = \frac{1}{( \frac{20 x ^{5} y ^{2}}{5 x ^{- 3} y ^{7}} ) ^{3}}

= \frac{1}{( \frac{20 x ^{5} y ^{2}}{5 x ^{- 3} y ^{7}} ) ^{3}}

Simplify

(\frac{20 x ^{5} y ^{2}}{5 x ^{- 3} y ^{7}})^{3} : \frac{64 x ^{24}}{y ^{15}}

= \frac{1}{\frac{64 x ^{24}}{y ^{15}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{y ^{15}}{64 x ^{24}}

\frac{5 ^{8}}{5 ^{5}}

3 n^{3} + 6 n^{2}

z + (\frac{1}{2} + \frac{1}{2} i) = i

3 x^{2} + 5 x - 10 = 0

(\frac{1}{81})^{\frac{3}{4}}

What is ((20x^5y^2)/(5x^{-3)y^7})^{-3} ?
The solution to ((20x^5y^2)/(5x^{-3)y^7})^{-3} is (y^{15})/(64x^{24)}