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Thread: Adding rational expr: 5 / (x^3 - y^3) + 3 / (x^2 + xy + y^2)

  1. #1
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    Adding rational expr: 5 / (x^3 - y^3) + 3 / (x^2 + xy + y^2)

    The problem: 5 / (x^3 - y^3) + 3 / (x^2 + xy + y^2) = ?

    My work so far: 5 / (x - y(x^2 - y^2)) + 3 / [(x + y)(x + y)]

    I don't know what to do from here for a common denominator

    thank you
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    Edited by stapel -- Reason for edit: attempting to restore formatting and thus meaning

  2. #2
    Elite Member
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    Is this your problem?. Try using LaTex. Your post got all mixed up. Hard to read.

    [tex]\L\\\frac{5}{x^{3}-y^{3}}+\frac{3}{x^{2}+xy+y^{2}}[/tex]

    If this is correct, you should recognize the difference of two cubes.

    [tex]\L\\x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})[/tex]

  3. #3
    Elite Member
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    Re: Adding rational expressions

    Quote Originally Posted by alee
    Code:
    5                  +                    3                             
    -------------               ------------------   =                                          
    x^3 - y^3      +         x^2 + xy + y^2

    Are these two separate problems?

    Code:
        5                    +              3
    ---------------          ---------------   = 
     x-y(x^2-y^2)    +     (x+y)(x+y)
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  4. #4
    Senior Member skeeter's Avatar
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    Re: Adding rational expressions

    Quote Originally Posted by alee
    5 + 3
    ------------- ------------------ =
    x^3 - y^3 + x^2 + xy + y^2



    5 + 3
    --------------- --------------- =
    x-y(x^2-y^2) + (x+y)(x+y)


    I don't know what to do from here for a common denominator

    thank you
    first, note that [tex]\L x^2 + xy + y^2 \neq (x+y)(x+y)[/tex].
    second, note galactus' factorization of [tex]\L x^3 - y^3[/tex]. the common denominator will be [tex]\L x^3 - y^3 = (x-y)(x^2 + xy + y^2)[/tex] ...

    [tex]\L \frac{5}{(x-y)(x^2 + xy + y^2)} + \frac{3(x-y)}{(x-y)(x^2 + xy + y^2)}[/tex] ... add'em up.

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