How do you simplify ${(\frac{3 x^{- 2} y^{3}}{2 x y})}^{- 2}$ $((3x^-2 y^3 )/(2xy))^-2$ ?

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How do you simplify ${(\frac{3 x^{- 2} y^{3}}{2 x y})}^{- 2}$ $((3x^-2 y^3 )/(2xy))^-2$ ?

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Answer 1 · 2015-10-11T00:08:43

$\frac{4 x^{6}}{9 y^{4}}$ $(4x^6)/(9y^4)$

Explanation:

Try to simply the expression

$\frac{3 x^{- 2} y^{3}}{2 x y}$ $(3x^(-2)y^3)/(2xy)$

first, then worry about the $- 2$ $-2$ exponent.

Notice that you can write

$\frac{3 x^{- 2} y^{3}}{2 x y} = \frac{3 \cdot y^{3}}{2 x y} \cdot \frac{1}{x^{2}} = \frac{3 y^{2}}{2 x^{3}}$ $(3x^(-2)y^3)/(2xy) = (3 * y^3)/(2xy) * 1/x^2 = (3y^2)/(2x^3)$

The original expression now becomes

${(\frac{3 y^{2}}{2 x^{3}})}^{- 2} = {(\frac{2 x^{3}}{3 y^{2}})}^{2} = \frac{2^{2} \cdot x^{3 \cdot 2}}{3^{2} \cdot y^{2 \cdot 2}} = \frac{4 x^{6}}{9 y^{4}}$ $((3y^2)/(2x^3))^(-2) = ((2x^3)/(3y^2))^2 = (2^2 * x^(3 * 2))/(3^2 * y^(2 * 2)) = color(green)((4x^6)/(9y^4))$

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How do you simplify ${(\frac{3 x^{- 2} y^{3}}{2 x y})}^{- 2}$ $((3x^-2 y^3 )/(2xy))^-2$ ?

1 Answer

Explanation:

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