What is ((x^{-3}y^4)^3)/((-3x^2y^{-2))^2} ?

The solution to ((x^{-3}y^4)^3)/((-3x^2y^{-2))^2} is (y^{16})/(9x^{13)}

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simplify ((x^{-3}y^4)^3)/((-3x^2y^{-2))^2}

simplify

\frac{( x ^{- 3} y ^{4} ) ^{3}}{( - 3 x ^{2} y ^{- 2} ) ^{2}}

\frac{y ^{16}}{9 x ^{13}}

Solution steps

\frac{( x ^{- 3} y ^{4} ) ^{3}}{( - 3 x ^{2} y ^{- 2} ) ^{2}}

Simplify

(x^{- 3} y^{4})^{3} : \frac{y ^{12}}{x ^{9}}

= \frac{\frac{y ^{12}}{x ^{9}}}{( - 3 x ^{2} y ^{- 2} ) ^{2}}

Apply the fraction rule:

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

= \frac{y ^{12}}{x ^{9} ( - 3 x ^{2} y ^{- 2} ) ^{2}}

Simplify

x^{9} (- 3 x^{2} y^{- 2})^{2} : \frac{9 x ^{13}}{y ^{4}}

= \frac{y ^{12}}{\frac{9 x ^{13}}{y ^{4}}}

Apply the fraction rule:

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

= \frac{y ^{12} y ^{4}}{9 x ^{13}}

Simplify

y^{12} y^{4} : y^{16}

= \frac{y ^{16}}{9 x ^{13}}

25 \cdot \frac{1}{2}

- 1^{15}

\frac{48}{49}

e^{2 x} \cdot (- e^{- x})

\frac{sec ( a )}{tan ( a )}

What is ((x^{-3}y^4)^3)/((-3x^2y^{-2))^2} ?
The solution to ((x^{-3}y^4)^3)/((-3x^2y^{-2))^2} is (y^{16})/(9x^{13)}