# How do you evaluate (2y ^ { 2} \cdot y ^ { 3} \cdot 3y ) ^ { 3}?

Nov 14, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

${\left(\left(2 \cdot 3\right) \left({y}^{2} \cdot {y}^{3} \cdot y\right)\right)}^{3} \implies$

${\left(6 \left({y}^{2} \cdot {y}^{3} \cdot y\right)\right)}^{3}$

Next, use these rules of exponents to evaluate the $y$ terms within the parenthesis:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

${\left(6 \left({y}^{2} \cdot {y}^{3} \cdot y\right)\right)}^{3} \implies$

${\left(6 \left({y}^{\textcolor{red}{2}} \cdot {y}^{\textcolor{b l u e}{3}} \cdot {y}^{\textcolor{p u r p \le}{1}}\right)\right)}^{3} \implies$

${\left(6 {y}^{\textcolor{red}{2} + \textcolor{b l u e}{3} + \textcolor{p u r p \le}{1}}\right)}^{3} \implies$

${\left(6 {y}^{6}\right)}^{3}$

Now, use these rules of exponents to complete the evaluation:

$a = {a}^{\textcolor{red}{1}}$ and ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left(6 {y}^{6}\right)}^{3} \implies$

${\left({6}^{\textcolor{red}{1}} {y}^{\textcolor{red}{6}}\right)}^{\textcolor{b l u e}{3}} \implies$

${6}^{\textcolor{red}{1} \times \textcolor{b l u e}{3}} {y}^{\textcolor{red}{6} \times \textcolor{b l u e}{3}} \implies$

${6}^{3} {y}^{18} \implies$

$216 {y}^{18}$