# How do you multiply (- 5k ^ { 2} j ^ { 4} ) ^ { 3} ( k ^ { 7} j ^ { 7} ) ^ { 3}?

Feb 4, 2017

See the entire solution process below:

#### Explanation:

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

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

${\left(- 5 {k}^{\textcolor{red}{2}} {j}^{\textcolor{red}{4}}\right)}^{\textcolor{b l u e}{3}} {\left({k}^{\textcolor{red}{7}} {j}^{\textcolor{red}{7}}\right)}^{\textcolor{b l u e}{3}} = \left(- {5}^{\textcolor{b l u e}{3}} {k}^{\textcolor{red}{2} \times \textcolor{b l u e}{3}} {j}^{\textcolor{red}{4} \times \textcolor{b l u e}{3}}\right) \left({k}^{\textcolor{red}{7} \times \textcolor{b l u e}{3}} {j}^{\textcolor{red}{7} \times \textcolor{b l u e}{3}}\right) =$

$\left(- 125 {k}^{6} {j}^{12}\right) \left({k}^{21} {j}^{21}\right)$

Next, rearrange the expression grouping like terms:

$- 125 \left({k}^{6} {k}^{21}\right) \left({j}^{12} {j}^{21}\right)$

We can now use this rule for exponents to complete the multiplication: ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$- 125 \left({k}^{\textcolor{red}{6}} \times {k}^{\textcolor{b l u e}{21}}\right) \left({j}^{\textcolor{red}{12}} \times {j}^{\textcolor{b l u e}{21}}\right) = - 125 {k}^{\textcolor{red}{6} + \textcolor{b l u e}{21}} {j}^{\textcolor{red}{12} + \textcolor{b l u e}{21}} =$

$- 125 {k}^{27} {j}^{33}$