# How do you simplify (2gh^4)^3[(-2g^4h)^3]^2?

May 28, 2017

$512 {g}^{27} {h}^{18}$

#### Explanation:

Start by taking the outside exponents and multiplying them through:

${\left(2 g {h}^{4}\right)}^{\textcolor{red}{3}} {\left[{\left(- 2 {g}^{4} h\right)}^{3}\right]}^{\textcolor{red}{2}}$

$= \left({2}^{\textcolor{red}{3}} {g}^{\textcolor{red}{3}} {h}^{4 \times \textcolor{red}{3}}\right) \left[{\left(- 2 {g}^{4} h\right)}^{3 \times \textcolor{red}{2}}\right]$

Simplify

$= \left(8 {g}^{3} {h}^{12}\right) \left[{\left(- 2 {g}^{4} h\right)}^{6}\right]$

Next, take the outside exponent of $6$ and multiply it through:

$= \left(8 {g}^{3} {h}^{12}\right) \left[{\left(- 2 {g}^{4} h\right)}^{\textcolor{b l u e}{6}}\right]$

$= \left(8 {g}^{3} {h}^{12}\right) \left({\left(- 2\right)}^{\textcolor{b l u e}{6}} {g}^{4 \times \textcolor{b l u e}{6}} {h}^{\textcolor{b l u e}{6}}\right)$

Simplify

$= \left(8 {g}^{3} {h}^{12}\right) \left({\left(- 2\right)}^{\textcolor{b l u e}{6}} {g}^{4 \times \textcolor{b l u e}{6}} {h}^{\textcolor{b l u e}{6}}\right)$

$= \left(8 {g}^{3} {h}^{12}\right) \left(64 {g}^{24} {h}^{6}\right)$

When multiplying two identical bases with different exponents, you add the exponents over a single base.

$= 8 \times 64 {g}^{3} {g}^{24} {h}^{12} {h}^{6}$

$= 512 {g}^{3 + 24} {h}^{12 + 6}$

$= 512 {g}^{27} {h}^{18}$