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Math Help - Simplifying question

  1. #1
    Newbie
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    Simplifying question

    Hello,

    I was wondering how to simplify the following expression:



    I hope the picture is clear enough and that someone can help me
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  2. #2
    Junior Member lanzailan's Avatar
    Joined
    Mar 2009
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    Hello Barbor..
    let me try....

    (1/6x)^(1/3).(4/6)^2/3
    =(1/6)^(1/3).(4/6)^(2/3).x^(1/3).x^(2/3)
    =(1/6)^(1/3).(4/6)^2(1/3).x^(1/3+2/3)
    =(1/6)^(1/3).(4/6)^(1/3).(4/6)^(1/3).x
    =(1/6)^(1/3).(4)^(1/3).(1/6)^(1/3).(4)^(1/3).(1/6)^(1/3).x
    =(1/6)^(1/3+1/3+1/3).4^(1/3).x
    =(1/6)^(3/3).4^(1/3).x
    =1/6.4^(1/3).x
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  3. #3
    Super Member

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    Hello, Barbor!


    \left(\frac{1}{6}\,x\right)^{\frac{1}{3}}\left(\fr  ac{4}{6}\,x\right)^{\frac{2}{3}}

    We have: . \left(\frac{1}{6}\right)^{\frac{1}{3}}\!\!\cdot x^{\frac{1}{3}} \cdot\!\left(\frac{4}{6}\right)^{\frac{2}{3}}\!\!\  cdot x^{\frac{2}{3}} \;\;=\;\;\frac{1^{\frac{1}{3}}}{6^{\frac{1}{3}}} \cdot\frac{4^{\frac{2}{3}}}{6^{\frac{2}{3}}} \cdot x^{\frac{1}{3}} \cdot x^{\frac{2}{3}} \;\;=\; \;\frac{4^{\frac{2}{3}}}{6^1} \cdot x^1

    . . . =\;\;\frac{(2^2)^{\frac{2}{3}}}{6}\,x \;\;=\;\; \frac{2^{\frac{4}{3}}}{2\cdot3}\,x \;\;=\;\; \frac{2^{\frac{1}{3}}}{3}\,x \;=\;\frac{\sqrt[3]{2}}{3}\,x

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  4. #4
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    Thank you very much!
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