Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: How to solve this?
  1. How to solve? (Replies: 1)

  2. How to solve (Replies: 9)

  3. :How to solve this? (Replies: 1)

  4. How to solve this? (Replies: 34)

  5. How to solve? (Replies: 1)