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Math Help - Rationalize the denominator

  1. #1
    Member purplec16's Avatar
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    Rationalize the denominator

    [\MATH]\frac{1}{((x)^(1/3)-(y)^(1/3))}[/MATH]
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  2. #2
    Member purplec16's Avatar
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    Re: Rationalize the denominator

    \frac{1}{((x)^(1/3)-(y)^(1/3))}
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  3. #3
    MHF Contributor Reckoner's Avatar
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    Re: Rationalize the denominator

    \frac1{x^{1/3}-y^{1/3}}

    =\frac1{x^{1/3}-y^{1/3}}\cdot\frac{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}

    =\frac{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}{x-y}

    Note that the expression which I multiplied the numerator and denominator by is just the second half of the difference of cubes formula.
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  4. #4
    Member purplec16's Avatar
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    Re: Rationalize the denominator

    I've noticed... but I don't understand why you have to do that but rules are rules I guess... Thank you so much for your response!!! But how is that x-y in the denominator?
    Last edited by purplec16; June 5th 2012 at 04:56 AM.
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  5. #5
    MHF Contributor Reckoner's Avatar
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    Re: Rationalize the denominator

    Quote Originally Posted by purplec16 View Post
    I've noticed... but I don't understand why you have to do that but rules are rules I guess... Thank you so much for your response!!! But how is that x-y in the denominator?
    \left(x^{1/3}-y^{1/3}\right)\left(x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}\right)

    Distributing, we have

    =x+x^{2/3}y^{1/3}+x^{1/3}y^{2/3}-x^{2/3}y^{1/3}-x^{1/3}y^{2/3}-y

    Now, combining like terms leaves

    =x-y.
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  6. #6
    Member purplec16's Avatar
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    Re: Rationalize the denominator

    Okay. Thanks for your help!
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