# How do you simplify (a^-1b^(1/3)*a^(-4/3)b^2)^2?

Jan 15, 2017

See entire simplification process below:

#### Explanation:

First, we can use this rule for exponents to start the simplification process:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left({a}^{\textcolor{red}{- 1}} {b}^{\textcolor{red}{\frac{1}{3}}} \cdot {a}^{\textcolor{red}{- \frac{4}{3}}} {b}^{\textcolor{red}{2}}\right)}^{\textcolor{b l u e}{2}} \to$

$\left({a}^{\textcolor{red}{- 1} \times \textcolor{b l u e}{2}} {b}^{\textcolor{red}{\frac{1}{3}} \times \textcolor{b l u e}{2}} \cdot {a}^{\textcolor{red}{- \frac{4}{3}} \times \textcolor{b l u e}{2}} {b}^{\textcolor{red}{2} \times \textcolor{b l u e}{2}}\right) \to$

${a}^{-} 2 {b}^{\frac{2}{3}} \cdot {a}^{- \frac{8}{3}} {b}^{4}$

Next, we can group like terms:

$\left({a}^{-} 2 \cdot {a}^{- \frac{8}{3}}\right) \left({b}^{\frac{2}{3}} \cdot {b}^{4}\right)$

Next, we can use this rule of exponents to further simplify:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\left({a}^{\textcolor{red}{- 2}} \cdot {a}^{\textcolor{b l u e}{- \frac{8}{3}}}\right) \left({b}^{\textcolor{red}{\frac{2}{3}}} \cdot {b}^{\textcolor{b l u e}{4}}\right)$

$\left({a}^{\textcolor{red}{- 2 \times \frac{3}{3}}} \cdot {a}^{\textcolor{b l u e}{- \frac{8}{3}}}\right) \left({b}^{\textcolor{red}{\frac{2}{3}}} \cdot {b}^{\textcolor{b l u e}{4 \times \frac{3}{3}}}\right)$

$\left({a}^{\textcolor{red}{- \frac{6}{3}}} \cdot {a}^{\textcolor{b l u e}{- \frac{8}{3}}}\right) \left({b}^{\textcolor{red}{\frac{2}{3}}} \cdot {b}^{\textcolor{b l u e}{\frac{12}{3}}}\right)$

$\left({a}^{\textcolor{red}{- \frac{6}{3}} + \textcolor{b l u e}{- \frac{8}{3}}}\right) \left({b}^{\textcolor{red}{\frac{2}{3}} + \textcolor{b l u e}{\frac{12}{3}}}\right)$

${a}^{- \frac{14}{3}} {b}^{\frac{14}{3}}$

We can now use this rule of exponents to further transform this expression:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

${b}^{\frac{14}{3}} / {a}^{- - \frac{14}{3}}$

${b}^{\frac{14}{3}} / {a}^{\frac{14}{3}}$

or

${\left(\frac{b}{a}\right)}^{\frac{14}{3}}$