# How do you multiply and simplify (\frac { 2x y ^ { 2} \cdot - 2x ^ { 0} y ^ { 2} } { - 2x ^ { - 1} y ^ { 0} } ) ^ { 3}?

Jun 26, 2018

The answer is $8 {x}^{6} {y}^{12}$.

#### Explanation:

${\left(\frac{2 x {y}^{2} \cdot \left(- 2 {x}^{0} {y}^{2}\right)}{- 2 {x}^{-} 1 {y}^{0}}\right)}^{3}$

First, remember that anything raised to the 0th power is equal to $1$. So start by canceling out all the terms that are raised to $0$.

${\left(\frac{2 x {y}^{2} \cdot \left(- 2 \left(1\right) {y}^{2}\right)}{- 2 {x}^{-} 1 \left(1\right)}\right)}^{3}$

${\left(\frac{2 x {y}^{2} \cdot \left(- 2 {y}^{2}\right)}{- 2 {x}^{-} 1}\right)}^{3}$

Multiply the numerator by adding the exponent of like terms. Don't forget the negative sign!

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

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

Remember that anything raised to the $- 1$ is basically a reciprocal, so:

${x}^{-} 1 = \frac{1}{x}$

Therefore:

${\left(\frac{2 x {y}^{4}}{\frac{1}{x}}\right)}^{3}$

which is

${\left(2 x {y}^{4} \cdot x\right)}^{3}$

${\left(2 {x}^{2} {y}^{4}\right)}^{3}$

Distribute the exponent

${2}^{3} \cdot {x}^{2 \cdot 3} \cdot {y}^{4 \cdot 3}$

$= 8 {x}^{6} {y}^{12}$