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How to solve this?

  1. Oct 27, 2003 #1
    Hi guyz,

    How can I solve the equation with the form :

    ax^3 + bx^2 + cx + d = 0

    I want the general way to solve allllllllll cubic equations..

  2. jcsd
  3. Oct 27, 2003 #2
  4. Oct 28, 2003 #3


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    The post above is very good but here it is in a nutshell:

    We know that (a- b)3= a3-3a2b+ 3ab2- b3 and that 3ab(a-b)= 3a2b- 3ab so:

    (a-b)3+ 3ab(a-b)= a3- b3
    (the middle terms cancel).

    That is: if we pick any two numbers a, b and let x= a- b, m= 3ab and n= a3- b3, then x3+ mx= n.

    What about the other way? If we are given m and n, can we solve for a and b (and so find x)?

    Yes, we can. From m= 3ab, we have b= m/(3a). Put that into
    a3- b3= n and we have a3- m3/(33a3)= n

    Multiply both sides of the equation by a3 and we have the (6th degree!!) equation
    a6- (m/3)3= naa3.

    But if we let u= a3, this reduces to a quadratic equation for u: u2- nu- (m/3)3= 0.

    We can solve that by the quadratic formula:
    u= a3= (n +/- √(n2+ 4(m/3)3))/2= (n/2)+/- √((n/2)2+ (m/3)3).

    Since a3- b3= n,
    b3= a3- n
    = (-n/2)+/- √((n/2)2+ (m/3)3).

    Solving for a and b,

    a= (((n/2)+/- √((n/2)2+ (m/3)3))1/3
    b= ((-(n/2)+/- √((n/2)2+ (m/3)3))1/3

    and, finally, x= a-b.

    Notice that the equation was x3+ mx- n= 0 which has no x2. This is the "reduced" cubic.

    To solve a general cubic, x3+ ax2+ bx+ c= 0,
    "Shift" the variable: replace x by y+ y0 in this equation and then choose y0 to make the coefficient of y2 equal to 0.
  5. Oct 29, 2003 #4
    Thanks Guyz,

    Thanks Guyz, I hope these equations are true, because i can't check them right now :)

    I'm very thankful :):)
  6. Oct 29, 2003 #5

    Hey Guyz, Does the equation "imhereyeah" posted solves with 3 values of x????? or 1 only?
  7. Nov 2, 2003 #6


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    Re: ?????

    Yes. Note that the "a" and "b" in that reply are derived from cube roots. So "a" and "b" each have 3 possible (complex) values giving a total of 9 combinations in the solution (though at most 3 will be distinct).
    Last edited: Nov 2, 2003
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