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Math Help - Complicated Fractions

  1. #1
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    Complicated Fractions

    Given a+b+c=0, evaluate \frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}+\frac{1}{a^2+b^2-c^2}.
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  2. #2
    Moo
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    Hello,

    Let see what a^2+b^2-c^2 can give :

    a^2+b^2-c^2=a^2+(b+c)(b-c)

    We know that a+b+c=0 <=> b+c=-a

    a^2+(b+c)(b-c)=a^2-a(b-c)=a(a-b+c)

    As a+c=-b, we finally have :

    a^2+b^2-c^2=-2ab

    Hence we get :

    \frac{1}{a^2+b^2-c^2}=\frac{1}{-2ab}

    With a similar method, we get :

    \frac{1}{a^2+c^2-b^2}=\frac{1}{-2ac}

    \frac{1}{b^2+c^2-a^2}=\frac{1}{-2bc}



    ===> \frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}+\frac{1}{a^2+b^2-c^2} = \frac{1}{-2ab}+\frac{1}{-2ac}+\frac{1}{-2bc}

    = \frac{c}{-2abc}+\frac{b}{-2abc}+\frac{a}{-2abc} = \frac{a+b+c}{-2abc} = 0
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