# What is the right answer?

Mar 3, 2018

C

#### Explanation:

Use the exponential rules

• ${\left({x}^{a}\right)}^{b} = {x}^{a b}$

• ${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

Therefore

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

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

$= {x}^{- \frac{2}{3}} \cdot {y}^{\frac{4}{3}} \cdot {x}^{2} \cdot {y}^{- \frac{1}{3}}$

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

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

$= {x}^{\frac{4}{3}} \cdot y$

Mar 3, 2018

I will let you do the final part

#### Explanation:

Given: $\textcolor{g r e e n}{{\left({x}^{- 2} {y}^{4}\right)}^{\frac{1}{3}} {x}^{2} {y}^{- \frac{1}{3}}}$

The trick is to take at least some of it one step at a time. Other parts you may be able to jump steps.

If you are unsure about the negative indices do this:

Write as color(green)( (y^4/x^2)^(1/3) x^2/y^(1/3)

color(green)( (y^(4/3)/x^(2/3)) x^2/y^(1/3)

$\textcolor{g r e e n}{\frac{{y}^{\frac{4}{3}}}{{y}^{\frac{1}{3}}} \times {x}^{2} / \left({x}^{\frac{2}{3}}\right)}$

$\textcolor{g r e e n}{{y}^{\frac{4}{3} - \frac{1}{3}} \textcolor{w h i t e}{.} {x}^{2 - \frac{2}{3}}}$

I will let you finish this off