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Cubing the Cubic Formula Help would be greatly appreciated

  1. Nov 27, 2011 #1
    Hi,

    So I was working on a little project the other day and it was in regards to the cubic formula...

    Which can be found here:

    http://en.wikipedia.org/wiki/Cubic_formula#General_formula_of_roots

    basically given an equation of the form:

    ax3 + bx2 + cx + d = 0 the formulas on the attached link will give you the three values of x that satisfy this equation.

    Here is my dilemma,

    I was just kind of playing around with the quadratic equation which is:

    x = [itex]\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/itex]

    It can be conveniently "re arranged" to look like this:

    (2ax + b)2 = b2 - 4ac

    So in dealing with the cubic equation I tried the same thing which was fairly simple:

    (3ax + b) = Ω (where Ω is all of our other business that remains)...

    Can somebody tell me the simplified value of Ω3?

    Its terribly difficult to solve by hand and I do not own a symbolic calculator can manipulate it so if somebody could do the solution that would be convenient...

    Additionally I would like to know another value:

    Ω can be seen as the sum of (σ + μ) (see the two giant cube roots in the wiki article, those are sig and mu for this expression).

    What is (σ2 - σμ + μ2)?

    and What is (σ3 - μ3)

    Hopefully between those three questions I can resolve to find a pattern between the quadratic and cubic formulas.

    Thanks!
     
  2. jcsd
  3. Nov 27, 2011 #2

    epenguin

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    I won't go through the detail of something that is in all the books just some overall ideas.

    Playing around is good. You have discovered a little thing you had probably forgotten (often happens). That actually if quadratic equation solving was explained to you, it goes in the reverse order - you don't get the second formula from the first, you get the first from the second, which is the quadratic to be solved rearranged. You try to incorporate its first two terms into a perfect square and then you have a constant, (b2 - 4ac) left over. So you can solve taking square roots.

    Put another way you get the difference of two squares

    (x + b/2)2 - {√[(b2 - 4ac)/4a2]}2 = 0

    which, taking a term to the other side and taking square roots leads to two linear equations.

    For the cubic you try to do the same idea (to answer your question) - you try to express it as the difference of two cubes, e.g. if we make the leading term 1, then try to express any cubic with this leading term as

    (x + β)3 - [(1 - α3)x3 + γ)3.

    When you expand and try to identify coefficients with those of a cubic you find you need to solve a quadratic equation.

    If the cubic has three real roots the roots of this quadratic are nonreal. It was the genius idea you could use them anyway and through them get the real roots of the cubic that was the takeoff that made math something else.

    You can do the quartic too as difference of two squares, one being the square of a square; that involves solving a cubic equation.

    Higher degree than that you can rearrange polynomials in pretty ways but this does into allow you to solve them, it is well known
     
  4. Nov 27, 2011 #3
    what is the proof that quintics and higher are unsolvabe? (i'm not familiar with galois theory or lie groups so i'm going to need it slightly more explained)
     
  5. Nov 27, 2011 #4

    gb7nash

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    Unfortunately, I don't think there is a proof that doesn't involve Galois theory.
     
  6. Nov 27, 2011 #5
    could u give me the proof (with theory) and i'll see if i can grasp something about it
     
  7. Nov 27, 2011 #6

    mathwonk

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    the best explanation i know of the cubic formula is in euler's elements of algebra. it is free on the web.
     
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