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Math Help - algebraic problem

  1. #1
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    algebraic problem

    Challenge problem:

    Express the following algebraic quantities in terms of the variables λ,μ :

    1) a^2+b^2+c^2

    2) a^3+b^3+c^3

    3) a^5+b^5+c^5


    Where  a\neq b\neq c are the roots of the following equation:

    x^3+\lambda x+\mu=0 , \lambda\neq 0,\mu\neq 0 and real Nos



    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; June 23rd 2010 at 04:16 PM. Reason: Approved.
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  2. #2
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    (a)  a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca) = -2A

    (b) we have  a^3 + A a + B = 0

     a^3 = -Aa  - B

     a^3 + b^3 + c^3 = -A(a+b+c) - 3B = -3B

    (c)  a^5 + Aa^3 + Ba^2 = 0

     a^5 = -Aa^3 - Ba^2

     a^5 + b^5 + c^5 = -A(a^3 + b^3 + c^3) - B ( a^2 + b^2 + c^2) = 3AB + 2AB = 5AB
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  3. #3
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    Quote Originally Posted by simplependulum View Post
    (a)  a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca) = -2A

    (b) we have  a^3 + A a + B = 0

     a^3 = -Aa  - B

     a^3 + b^3 + c^3 = -A(a+b+c) - 3B = -3B

    (c)  a^5 + Aa^3 + Ba^2 = 0

     a^5 = -Aa^3 - Ba^2

     a^5 + b^5 + c^5 = -A(a^3 + b^3 + c^3) - B ( a^2 + b^2 + c^2) = 3AB + 2AB = 5AB

    O.K correct ,but why don't you show how you get that :

    a+b+c=0 and ab+ac+bc = 2A ? Not for me but for other people that want to know the solution of the problem?
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