# How do you simplify 2a^2(4a^-2b^3)^-3 and write it using only positive exponents?

Jan 26, 2017

See the entire simplification process below:

#### Explanation:

First, expand the term in parenthesis using this rule for exponents:

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

$2 {a}^{2} {\left(4 {a}^{-} 2 {b}^{3}\right)}^{-} 3 \to 2 {a}^{2} \left({4}^{-} 3 {a}^{- 2 \times - 3} {b}^{3 \times - 3}\right) \to 2 {a}^{2} \left({4}^{-} 3 {a}^{6} {b}^{-} 9\right)$

Now, group like terms:

$\left(2 \times {4}^{-} 3\right) \left({a}^{2} \times {a}^{6}\right) {b}^{-} 9$

Then, combine like terms using this rule of exponents:

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

Giving:

$2 {a}^{2 + 6} {4}^{-} 3 {b}^{-} 9 \to 2 {a}^{8} {4}^{-} 3 {b}^{-} 9$

Now, use this rule of exponents to remove the negative exponents:

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

Giving:

$\frac{2 {a}^{8}}{{4}^{- - 3} {b}^{- - 9}} \to \frac{2 {a}^{8}}{{4}^{3} {b}^{9}} \to \frac{2 {a}^{8}}{64 {b}^{9}} \to$

${a}^{8} / \left(32 {b}^{9}\right)$