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Thread: Rationalizing denominators with cube roots

  1. #1
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    Rationalizing denominators with cube roots

    $\displaystyle \sqrt[3]{5} $
    ____
    $\displaystyle \sqrt[3]{10} $
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  2. #2
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    Quote Originally Posted by Tenskypoo View Post
    $\displaystyle \sqrt[3]{5} $
    ____
    $\displaystyle \sqrt[3]{10} $
    Hi

    $\displaystyle \frac{\sqrt[3]{5}}{\sqrt[3]{10}} = \frac{5^{\frac{1}{3}}}{10^{\frac{1}{3}}} = \left(\frac{5}{10}\right)^{\frac{1}{3}} = \frac{1}{2^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{2}} = \frac{2^{\frac{2}{3}}}{2} = \frac{\sqrt[3]{4}}{2}$
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  3. #3
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    Quote Originally Posted by Tenskypoo View Post
    $\displaystyle \sqrt[3]{5} $
    ____
    $\displaystyle \sqrt[3]{10} $
    Cube root is a "third" of the number so you need two more:
    $\displaystyle \sqrt[3]{10}\sqrt[3]{10}\sqrt[3]{10}= \left(\sqrt[3]{10}\right)^3= 10$
    Multiply both numerator and denominator by $\displaystyle \left(\sqrt[3]{10}\right)^2$.
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