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Cardan's cubic equation

  1. Sep 6, 2008 #1
    [tex]X^3+PX+Q=0[/tex]

    [tex]X_0 = A + B [/tex]

    [tex](A+B)^3 + P(A+B) + Q = 0[/tex]

    [tex]A^3 + B^3 + (3AB+P)(A+B) + Q = 0[/tex]

    The next step is 3AB+P=0

    What make Cardano drop 3AB+P?

    Why it is zero?
     
  2. jcsd
  3. Sep 6, 2008 #2

    tiny-tim

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    Hi Physicsissuef! :smile:

    AB can be anything we like, so long as A + B = X.

    So we choose a ratio such that AB = -P/3.

    That eliminates one of the terms in the equation, and makes it much easier … a quadratic equation (in A^3, I think).

    And we already know how to solve a quadratic! :wink:
     
  4. Sep 7, 2008 #3
    Ok, thank you very much Timmy.
     
  5. Oct 17, 2010 #4
    could someone just show me with an example on how to work the cardano and Del ferro method from start to finish say using
    x^3-6x^2+2x-1
     
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