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Math Help - Square expresion

  1. #1
    Junior Member
    Joined
    Oct 2005
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    Square expresion

    It seems that the part of my last post that stated my tight sleep as true was indeed a lie. I now have another problem witch will haunt my dreems into infinity, that is iff someone doesn't help me a little!

    The problem is:
    \sqrt[3]{81}-\sqrt[6]{9}+\sqrt{12}-\sqrt[6]{27}-\sqrt[3]{3\sqrt3}
    Simplify this as far as possible.

    As for my attempt of a solution:
    \sqrt[3]{81}=\sqrt[3]{9*9}=\sqrt[3]{3*3*3*3}=3*\sqrt[3]{3}

    \sqrt[6]{9}=((3*3)^{1/2})^{1/3}=\sqrt[3]{3}

    \sqrt{12}=\sqrt{2*2*3}=2*\sqrt{3}

    \sqrt[6]{27}=\sqrt[6]{3*3*3}=((3*3*3)^{1/3})^{1/2}=\sqrt{3}

    \sqrt[3]{3*\sqrt{3}}=(3*\sqrt{3})^{1/3}=(3*3^{1/3})^{1/3}=3^{4/9}

    Thus the original expression equals:
    3*\sqrt[3]{3}-\sqrt[3]{3}+2\sqrt{3}-\sqrt{3}-3^{4/9} = }
    2*\sqrt[3]{3}+\sqrt{3}-3^{4/9} \not= 2*\sqrt[3]{3}

    Where 2*\sqrt[3]{3} is the answer given by my textbook.
    What did I do wrong?
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by a4swe
    \sqrt[3]{3*\sqrt{3}}=(3*\sqrt{3})^{1/3}=(3*3^{1/3})^{1/3}=3^{4/9}
    This is where you're off, it should be:

    \sqrt[3]{3*\sqrt{3}}=(3*\sqrt{3})^{1/3}=(3*3^{1/2})^{1/3}=3^{3/6}=\sqrt{3}

    Now let's finish: \sqrt[3]{81}-\sqrt[6]{9}+\sqrt{12}-\sqrt[6]{27}-\sqrt[3]{3\sqrt3}

    simplify: 3*\sqrt[3]{3}-\sqrt[3]{3}+2\sqrt{3}-\sqrt{3}-\sqrt3

    group like terms: \left(3*\sqrt[3]{3}-\sqrt[3]{3}\right)+\left(2\sqrt{3}-\sqrt{3}-\sqrt3\right)

    simplify: \left((3-1)*\sqrt[3]{3}\right)+\left(0\right)

    therefore: \boxed{2*\sqrt[3]{3}}
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  3. #3
    Junior Member
    Joined
    Oct 2005
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    Thank you very much, that was quick.
    I thought it was some minor error just like that witch I just couldn't spot.
    And belive me, I have looked for it, for a long time.
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