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Math Help - #3 Simplifying Radicals

  1. #1
    Member Jonboy's Avatar
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    #3 Simplifying Radicals

    Hello again; this two are the last probs for some time. As usual, please check my work.

    The directions are: Rationalize the denominator.

    \frac{1\,-\,\sqrt{2}}{2\sqrt{3}\,-\,\sqrt{6}}

    So I multilply by the conjugate & simplify: \frac{\sqrt{6}\,-\,2\sqrt{6}}{12\,-\,6}\;=\boxed{\;-\,\frac{\,\sqrt{6}}{6}}

    Now this beastly problem w/the same directions: \sqrt[3]{\frac{16}{9}}

    I thought to write it like this: \frac{\sqrt[3]{16}}{\sqrt[3]{9}}

    Then multiply the top and bottom by \sqrt[3]{3}: \frac{\sqrt[3]{48}}{3}\,=\,\boxed{\frac{2\sqrt[3]{6}}{3}}

    Does this look good? If you have another way that might be better, do say. Thanks once again!
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  2. #2
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    [size=3]Hello, Jonboy!

    Rationalize the denominator: . \frac{1\,-\,\sqrt{2}}{2\sqrt{3}\,-\,\sqrt{6}}

    So I multilply by the conjugate & simplify: \frac{\sqrt{6}\,-\,2\sqrt{6}}{12\,-\,6}\;=\boxed{\;-\,\frac{\,\sqrt{6}}{6}} . . . . Right!


    Now this beastly problem w/the same directions: \sqrt[3]{\frac{16}{9}}

    I thought to write it like this: \frac{\sqrt[3]{16}}{\sqrt[3]{9}}

    Then multiply the top and bottom by \sqrt[3]{3}: \frac{\sqrt[3]{48}}{3}\,=\,\boxed{\frac{2\sqrt[3]{6}}{3}} . . . . Great!
    You did this one in the most efficient way possible.

    Many (most?) would multiply top and bottom by \sqrt[3]{81}
    . . I did ... many years ago.

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  3. #3
    Member Jonboy's Avatar
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    Thanks for the confirmation Soroban! Also for posting that reply a pretty long time ago showing the different ways to rationalize the denominator, that's where I learned my shortcut.
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