What is ((3m^{-1}n^2)^4)/((2m^{-2)n)^3} ?

The solution to ((3m^{-1}n^2)^4)/((2m^{-2)n)^3} is (81n^5m^2)/8

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simplify ((3m^{-1}n^2)^4)/((2m^{-2)n)^3}

simplify

\frac{( 3 m ^{- 1} n ^{2} ) ^{4}}{( 2 m ^{- 2} n ) ^{3}}

\frac{81 n ^{5} m ^{2}}{8}

Solution steps

\frac{( 3 m ^{- 1} n ^{2} ) ^{4}}{( 2 m ^{- 2} n ) ^{3}}

Simplify

(3 m^{- 1} n^{2})^{4} : \frac{81 n ^{8}}{m ^{4}}

= \frac{\frac{81 n ^{8}}{m ^{4}}}{( 2 m ^{- 2} n ) ^{3}}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{81 n ^{8}}{m ^{4} ( 2 m ^{- 2} n ) ^{3}}

Simplify

m^{4} (2 m^{- 2} n)^{3} : \frac{8 n ^{3}}{m ^{2}}

= \frac{81 n ^{8}}{\frac{8 n ^{3}}{m ^{2}}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{81 n ^{8} m ^{2}}{8 n ^{3}}

Cancel

\frac{81 n ^{8} m ^{2}}{8 n ^{3}} : \frac{81 n ^{5} m ^{2}}{8}

= \frac{81 n ^{5} m ^{2}}{8}

- 18 = \frac{1}{4} y

p^{2} - 6 p + 8 = 0

(m + 2) (m + 3) = (m + 2) (m - 2)

\frac{x + 1}{x - 2} < 2

(x - 5) (x + 4) \geq 0

What is ((3m^{-1}n^2)^4)/((2m^{-2)n)^3} ?
The solution to ((3m^{-1}n^2)^4)/((2m^{-2)n)^3} is (81n^5m^2)/8