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Math Help - Adding a Rational Expression with difference of cubes.

  1. #1
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    Adding a Rational Expression with difference of cubes.

    I've been stuck on this question for about a half hour. I really don't think the answer I'm getting is correct.

    \frac {(x-y)^2}{x^3-y^3} + \frac {2}{x^2+xy+y^2}

    Is the LCD: (x-y)(x^2+xy+y^2)? I feel like I'm getting thrown off because I'm so use to factoring trinomials when finding the LCD.



    The answer I got was and it just doesn't look right:

    \frac {x^2-2xy+y^2+2x-2y}{(x-y)(x^2+xy+y^2)}
    Last edited by Latvija13; March 30th 2010 at 04:41 PM. Reason: Edited for Latex symbols
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  2. #2
    Senior Member Stroodle's Avatar
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    The numerator in your answer can be factorised to 2(x-y)+(x-y)^2

    So you can simplify the fraction to \frac{x-y+2}{x^2+xy+y^2}
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  3. #3
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    Quote Originally Posted by Stroodle View Post
    The numerator in your answer can be factorised to 2(x-y)+(x-y)^2

    So you can simplify the fraction to \frac{x-y+2}{x^2+xy+y^2}
    Thanks, I had forgotten that I could factor out an x-y from (x-y)^2 there. My teachers try so much to ingrain in your mind that you can't factor out things when there's a +/- sign separating expressions.
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  4. #4
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    Actually, it would have been simpler if you had recognized at the start that (x- y)(x^2+ xy+ y^2)= x^3- y^3.

    That is, you can immediately reduct the first fraction: \frac{(x- y)^2}{x^3- y^3}= \frac{(x- y)^2}{(x- y)(x^2+ xy+ y^2}= \frac{x- y}{x^2+ xy+ y^2}.

    Now, \frac{(x- y)^2}{x^3- y^3}+ \frac{2}{x^2+ xy+ y^2}= \frac{x-y}{x^2+ xy+ y^2}+ \frac{x- y}{x^2+ xy+ y^2}+ \frac{2}{x^2+ xy+ y^2} = \frac{x- y+ 2}{x^2+ xy+ y^2}.
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