# How do you rationalize the denominator and simplify root3((3a^8)/(4b))?

May 19, 2017

$\frac{\sqrt[3]{6 {a}^{8} {b}^{2}}}{2 b}$

#### Explanation:

We have: $\sqrt[3]{\frac{3 {a}^{8}}{4 b}}$

$= \frac{\sqrt[3]{3 {a}^{8}}}{\sqrt[3]{4 b}}$

$= \frac{\sqrt[3]{3 {a}^{8}}}{\sqrt[3]{4 b}} \cdot \frac{\sqrt[3]{4 b}}{\sqrt[3]{4 b}} \cdot \frac{\sqrt[3]{4 b}}{\sqrt[3]{4 b}}$

$= \frac{\sqrt[3]{48 {a}^{8} {b}^{2}}}{4 b}$

$= \frac{\sqrt[3]{48} \cdot \sqrt[3]{{a}^{8}} \cdot \sqrt[3]{{b}^{2}}}{4 b}$

$= \frac{\sqrt[3]{2 \cdot 2 \cdot 2 \cdot 6} \cdot \sqrt[3]{{a}^{8}} \cdot \sqrt[3]{{b}^{2}}}{4 b}$

$= \frac{\sqrt[3]{2 \cdot 2 \cdot 2} \cdot \sqrt[3]{6} \cdot \sqrt[3]{{a}^{8}} \cdot \sqrt[3]{{b}^{2}}}{4 b}$

$= \frac{2 \sqrt[3]{6 {a}^{8} {b}^{2}}}{4 b}$

$= \frac{\sqrt[3]{6 {a}^{8} {b}^{2}}}{2 b}$

May 19, 2017

color(blue)((root3(6a^8b^2))/(2b)

#### Explanation:

root3((3a^8)/(4b)

$\therefore = {\left(\frac{3 {a}^{8}}{4 b}\right)}^{\frac{1}{3}}$

$\therefore = \frac{{3}^{\frac{1}{3}} {a}^{\frac{8}{3}}}{{4}^{\frac{1}{3}} {b}^{\frac{1}{3}}}$

$\therefore = \frac{\sqrt[3]{3} \sqrt[3]{{a}^{8}}}{\sqrt[3]{4} \sqrt[3]{b}}$

$\therefore = \frac{\sqrt[3]{3 {a}^{8}}}{\sqrt[3]{4 b}}$

$\therefore = \frac{\sqrt[3]{3 {a}^{8}}}{\sqrt[3]{4 b}} \times \frac{\sqrt[3]{4 b}}{\sqrt[3]{4 b}} \times \frac{\sqrt[3]{4 b}}{\sqrt[3]{4 b}}$

$\therefore = \frac{\sqrt[3]{3 {a}^{8} \cdot 4 b \cdot 4 b}}{4 b}$

$\therefore = \sqrt[3]{4 b} \times \sqrt[3]{4 b} \times \sqrt[3]{4 b} = 4 b$

$\therefore = \frac{\sqrt[3]{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 {a}^{8} {b}^{2}}}{4 b}$

$\therefore = \sqrt[3]{2} \times \sqrt[3]{2} \times \sqrt[3]{2} = 2$

$\therefore = \frac{{\cancel{2}}^{1} \sqrt[3]{6 {a}^{8} {b}^{2}}}{{\cancel{4}}^{2} b}$

:.=color(blue)(root3(6a^8b^2)/(2b)