# How do you simplify ((2x^3y^2) /( 3xy))^ -3?

Mar 25, 2016

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

#### Explanation:

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

${\left(\frac{2 {x}^{3} {y}^{2}}{3 x y}\right)}^{-} 3 = {\left(\frac{3 x y}{2 {x}^{3} {y}^{2}}\right)}^{3}$

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

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

Distribute the third power exponent:

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

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

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

Cancel out like terms from top and bottom:

$\frac{{3}^{3}}{{2}^{3} {x}^{3} {y}^{2}}$

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

Mar 26, 2016

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

#### Explanation:

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

Example:$\text{ } \frac{1}{x} ^ 2$ can be written as $\text{ } {x}^{- 2}$

So we have$\text{ } \frac{2}{3} \times {x}^{3} {y}^{2} \times {x}^{- 1} {y}^{- 1}$

This gives us:$\text{ } \frac{2}{3} \times {x}^{3 - 1} {y}^{2 - 1}$

$\frac{2}{3} \times {x}^{2} y = \frac{2 {x}^{2} y}{3}$

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

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

This gives us:$\text{ } {\left(\frac{3}{2 {x}^{2} y}\right)}^{3}$

${3}^{3} = 27$

${2}^{3} = 8$

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

${\left(y\right)}^{3} = {y}^{3}$

Putting it all together

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