# How do you simplify  ((8b^4)/(2c^9))^3?

Apr 12, 2016

${\left(\frac{8 {b}^{4}}{2 {c}^{9}}\right)}^{3}$ can be simplified to $\frac{64 {b}^{12}}{c} ^ 27$

#### Explanation:

First, solve what you can inside the parentheses, then start using the other exponentiation:

${\left(\frac{8 {b}^{4}}{2 {c}^{9}}\right)}^{3}$ but $\frac{8}{2} = \frac{4}{1}$ so

${\left(\frac{8 {b}^{4}}{2 {c}^{9}}\right)}^{3} = {\left(\frac{4 {b}^{4}}{{c}^{9}}\right)}^{3}$ and then apply the exponentiation

${\left(\frac{4 {b}^{4}}{{c}^{9}}\right)}^{3} = \left(\frac{{4}^{1 \cdot 3} {b}^{4 \cdot 3}}{c} ^ \left(9 \cdot 3\right)\right)$
${4}^{3} = 64$, $1 \cdot 3 = 3$, $4 \cdot 3 = 12$ and $9 \cdot 3 = 27$ so
${\left(\frac{8 {b}^{4}}{2 {c}^{9}}\right)}^{3} = \frac{64 {b}^{12}}{c} ^ 27$