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

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

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Answer 1 · 2016-03-25T01:33:24

$\frac{27}{8 x^{3} y^{2}}$ $27/(8x^3y^2)$

Explanation:

The equation is raised to the negative third power, flip the fraction to turn it to a positive third power:

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

Then raise the numerator and the denominator by the third power:

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

Distribute the third power exponent:

$\frac{3^{3} x^{3} y^{3}}{2^{3} x^{9} y^{6}}$ $(3^3x^3y^3)/(2^3x^9y^6)$

Factor an $x^{3} y^{3}$ $x^3y^3$ from top and bottom:

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

Cancel out like terms from top and bottom:

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

Simplify: $\frac{27}{8 x^{3} y^{2}}$ $27/(8x^3y^2)$

Answer 2 · 2016-03-26T22:16:15

$\frac{27}{8 x^{6} y^{3}}$ $27/(8x^6y^3)$

Explanation:

For a moment let us disregard the power outside the brackets.

Example: $\frac{1}{x^{2}}$ $" " 1/x^2$ can be written as $x^{- 2}$ $" "x^(-2)$

So we have $\frac{2}{3} \times x^{3} y^{2} \times x^{- 1} y^{- 1}$ $" " 2/3xxx^3y^2xxx^(-1)y^(-1)$

This gives us: $\frac{2}{3} \times x^{3 - 1} y^{2 - 1}$ $" "2/3xx x^(3-1)y^(2-1)$

$\frac{2}{3} \times x^{2} y = \frac{2 x^{2} y}{3}$ $2/3xxx^2y = (2x^2y)/3$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Example: Suppose you have ${(\frac{1}{x})}^{- 3}$ $" " (1/x)^(-3)$ then this is ${(\frac{x}{1})}^{3}$ $" "(x/1)^3$

Ok, so your question has: ${(\frac{2 x^{2} y}{3})}^{- 3}$ $" "((2x^2y)/3)^(-3)$

This gives us: ${(\frac{3}{2 x^{2} y})}^{3}$ $" "(3/(2x^2y))^3$

$3^{3} = 27$ $3^3= 27$

$2^{3} = 8$ $2^3=8$

${(x^{2})}^{3} = x^{2 \times 3} = x^{6}$ $(x^2)^3 =x^(2xx3) = x^6$

${(y)}^{3} = y^{3}$ $(y)^3 = y^3$

Putting it all together

$\frac{27}{8 x^{6} y^{3}}$ $27/(8x^6y^3)$

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

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