How do you simplify root3 { 48b ^ { 10} }?

Nov 8, 2016

$2 {b}^{3} \sqrt[3]{6 b}$

Explanation:

Simplify $\sqrt[3]{48 {b}^{10}}$

Find factors of $48$ that are perfect cubes
(the cube of 2 is ${2}^{3} = 8$, the cube of 3 is ${3}^{3} = 27$, etc).

$48 = 8 \cdot 6 = {2}^{3} \cdot 6$

The cube root of ${2}^{3}$ is $2$. The${2}^{3}$ comes "out" of the cube root symbol as a $2$, while the $6$ stays inside.

$\sqrt[3]{48 {b}^{10}} = \sqrt[3]{8 \cdot 6 {b}^{10}} = \sqrt[3]{\textcolor{red}{{2}^{3}} \cdot 6 {b}^{10}} = \textcolor{red}{2} \sqrt[3]{6 {b}^{10}}$

Next lets look at the cube root of ${b}^{10}$.

If you divide $10$ by $3$, the result is $3$ with a remainder of $1$.
In other words ${b}^{10} = {b}^{3} \cdot {b}^{3} \cdot {b}^{3} \cdot {b}^{1}$. Each ${b}^{3}$ is a perfect cube, and comes "out" of the cube root symbol as $b$.

$2 \sqrt[3]{6 {b}^{10}} = 2 \sqrt[3]{6 \textcolor{red}{{b}^{3}} \cdot \textcolor{b l u e}{{b}^{3}} \cdot \textcolor{m a \ge n t a}{{b}^{3}} \cdot {b}^{1}} = 2 \textcolor{red}{b} \cdot \textcolor{b l u e}{b} \cdot \textcolor{m a \ge n t a}{b} \sqrt[3]{6 b}$

Multiplying the three $b$'s outside the cube root symbol gives

$2 {b}^{3} \sqrt[3]{6 b}$