What is ((x^3y^2)^{3/2})/((x^{-1)y^{-2/3})^{1/4}} ?

The solution to ((x^3y^2)^{3/2})/((x^{-1)y^{-2/3})^{1/4}} is x^{19/4}y^{19/6}

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simplify ((x^3y^2)^{3/2})/((x^{-1)y^{-2/3})^{1/4}}

simplify

\frac{( x ^{3} y ^{2} ) ^{\frac{3}{2}}}{( x ^{- 1} y ^{- \frac{2}{3}} ) ^{\frac{1}{4}}}

x^{\frac{19}{4}} y^{\frac{19}{6}}

Solution steps

\frac{( x ^{3} y ^{2} ) ^{\frac{3}{2}}}{( x ^{- 1} y ^{- \frac{2}{3}} ) ^{\frac{1}{4}}}

Simplify

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

= \frac{x ^{\frac{9}{2}} y ^{3}}{( x ^{- 1} y ^{- \frac{2}{3}} ) ^{\frac{1}{4}}}

Simplify

(x^{- 1} y^{- \frac{2}{3}})^{\frac{1}{4}} : \frac{1}{x ^{\frac{1}{4}} y ^{\frac{1}{6}}}

= \frac{x ^{\frac{9}{2}} y ^{3}}{\frac{1}{x ^{\frac{1}{4}} y ^{\frac{1}{6}}}}

Apply the fraction rule:

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

= \frac{x ^{\frac{9}{2}} y ^{3} x ^{\frac{1}{4}} y ^{\frac{1}{6}}}{1}

Cancel

\frac{x ^{\frac{9}{2}} y ^{3} x ^{\frac{1}{4}} y ^{\frac{1}{6}}}{1} : x^{\frac{19}{4}} y^{\frac{19}{6}}

= x^{\frac{19}{4}} y^{\frac{19}{6}}

\frac{2 ( x - 3 )}{3} = \frac{x + 1}{2}

6 (x + 1) + 2 \geq 5 x + 12

x^{2} + 10 x + 25

- 775 - 147

\frac{1}{2} (6)

What is ((x^3y^2)^{3/2})/((x^{-1)y^{-2/3})^{1/4}} ?
The solution to ((x^3y^2)^{3/2})/((x^{-1)y^{-2/3})^{1/4}} is x^{19/4}y^{19/6}