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Math Help - Tricky Problem

  1. #1
    Senior Member
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    Tricky Problem

    Ok so you are given; a, b, c are integers

    abc = 729
    and a + b + c = 91

    Find a^2 + b^2 + c^2

    How could I do this algebraically?
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  2. #2
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    Hint: 1+9+81=91
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  3. #3
    Super Member Bacterius's Avatar
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    Reply

    Say c = 1. Therefore, we have :

    ab = 729
    a + b = 90

    Substitute :

    a(90 - a) = 729
    90a - a^2 = 729
    a^2 - 90a + 729 = 0

    Solve for a.

    Facts for the lazy :
    Spoiler:
    You sure ?
    Spoiler:

    \triangle = 90^2 - 2916 = 5184

    a = \frac{\ {90 - \sqrt{\triangle}}}{2} = 9

    b = \frac{\ {90 + \sqrt{\triangle}}}{2} = 81

    And for the curious :
    Spoiler:
    Since you would get a polynomial with the same coefficients if you had substituted b instead of a, the two solutions of the quadratic equation are a and b.
    Oh, and by the way, you just factored 729 into smaller factors


    Finding the sum of the squares shouldn't be too hard now (remember c = 1)
    Last edited by Bacterius; November 6th 2009 at 07:31 PM.
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