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Math Help - cubic discriminant

  1. #1
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    cubic discriminant

    Hi guys, need help with this proof, have spent a while on it and getting no where.
    I followed the hint but couldn't do anything useful

    Suppose that the cubic polynomial x^3 + Ax + B factors into (x-a)(x-b)(x-c)
    Prove that 4A^3 + 27B^2 = 0 if and only if two (or more) of a, b, c are the same. (Hint. Multiply out the right-hand side and compare coefficients to relate A and B to a,b,c)

    Cheers for the help!
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: cubic discriminant

    Quote Originally Posted by liedora View Post
    Hi guys, need help with this proof, have spent a while on it and getting no where.
    I followed the hint but couldn't do anything useful

    Suppose that the cubic polynomial x^3 + Ax + B factors into (x-a)(x-b)(x-c)
    Prove that 4A^3 + 27B^2 = 0 if and only if two (or more) of a, b, c are the same. (Hint. Multiply out the right-hand side and compare coefficients to relate A and B to a,b,c)

    Cheers for the help!
    (x-a)(x-b)(x-c)=(x^2-bx-ax+ab)(x-c)=x^3-bx^2-ax^2+acx-cx^2+cbx+acx-abc

    x^3+Ax+B=x^3-bx^2-ax^2+acx-cx^2+cbx+acx-abc

    Hence,

    0=(a+b+c)

    A=ab+bc+ac

    B=-abc


    Now, what happens if a=b=c?  a=b? a=c?  b=c?
    Last edited by Also sprach Zarathustra; July 7th 2011 at 05:56 PM.
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  3. #3
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    Re: cubic discriminant

    Hi, I got that far and then didn't know where to go, I tried substituting those values into 4A^4 + 27B^2 but I didn't know what to do from there. Or if that is even the right step to take. I must be missing something!
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