# How do you simplify (9y)(2y^3)^2?

Feb 1, 2017

See the entire simplification process below:

#### Explanation:

Start the simplification process by using these rule for exponents:

$a = {a}^{\textcolor{red}{1}}$

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

$\left(9 y\right) {\left({2}^{\textcolor{red}{1}} {y}^{\textcolor{red}{3}}\right)}^{\textcolor{b l u e}{2}} \to \left(9 y\right) \left({2}^{\textcolor{red}{1} \times \textcolor{b l u e}{2}} {y}^{\textcolor{red}{3} \times \textcolor{b l u e}{2}}\right) \to \left(9 y\right) \left({2}^{2} {y}^{6}\right) \to \left(9 y\right) \left(4 {y}^{6}\right)$

Now, regroup like terms and use these rule of exponents to complete the simplification:

$a = {a}^{\textcolor{red}{1}}$

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

$\left(9 y\right) \left(4 {y}^{6}\right) \to \left(9 \times 4\right) \left(y \times {y}^{6}\right) \to 36 \left({y}^{\textcolor{red}{1}} \times {y}^{\textcolor{b l u e}{6}}\right) \to$

$36 \left({y}^{\textcolor{red}{1} + \textcolor{b l u e}{6}}\right) \to$

$36 {y}^{7}$