Rationalizing denominators with cube roots

• Apr 22nd 2009, 11:10 AM
Tenskypoo
Rationalizing denominators with cube roots
$\sqrt[3]{5}$
____
$\sqrt[3]{10}$
• Apr 22nd 2009, 11:30 AM
running-gag
Quote:

Originally Posted by Tenskypoo
$\sqrt[3]{5}$
____
$\sqrt[3]{10}$

Hi

$\frac{\sqrt[3]{5}}{\sqrt[3]{10}} = \frac{5^{\frac{1}{3}}}{10^{\frac{1}{3}}} = \left(\frac{5}{10}\right)^{\frac{1}{3}} = \frac{1}{2^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{2}} = \frac{2^{\frac{2}{3}}}{2} = \frac{\sqrt[3]{4}}{2}$
• Apr 22nd 2009, 11:45 AM
HallsofIvy
Quote:

Originally Posted by Tenskypoo
$\sqrt[3]{5}$
____
$\sqrt[3]{10}$

Cube root is a "third" of the number so you need two more:
$\sqrt[3]{10}\sqrt[3]{10}\sqrt[3]{10}= \left(\sqrt[3]{10}\right)^3= 10$
Multiply both numerator and denominator by $\left(\sqrt[3]{10}\right)^2$.