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Understanding Cardano's method of solving Cubic equation

  1. Dec 27, 2014 #1
    Dear All,
    I am trying to understanding Cardano's method to solve the cubic equation. Please first see the reference text that is attached (Alan F. Beardon, Algebra and Geometry).

    My rough understanding is that we make 2 substitutions (## P1(z - a/3)## and ##P(z - b/z) ## ) to simplify the cubic equation to the from (## z^6 -cz^3 - b^3 = 0 ##) so that we can make a further substitution (## y = z^3##). This results in a quadratic equation which can be solved to get two solutions y1 and y2.

    ## z^3 = y1## or ##y2 ## say y1 and y2 are 8 and -8 then this results in only two solutions right 2 and -2. Without forgetting that we need to do 2 Reverse substitutions to get the final solution. Why does the book say 6 solutions for z and only 3 are distinct?

    In the paragraph below from the text.
    " However, the values of ζ are the
    roots of the equation ## z^6 − cz^3 − b^3 = 0## , and if v is a root of this equation,
    then so (trivially) is v1 = −b/v. As v1 − b/v1 = v − b/v, we see that these six
    values of ζ can only provide at most three distinct roots of p. "

    what is v and v1. why is v1 = -b/v (The substitution was ##P(z - b/z) ## ) ??

    I think the coefficients, a, b & c from the equation 3.6.1 and the equation below 3.6.1 in the text are NOT the same as the original cubic equation (## z^3 + az^2 + bz + c = 0 ##). I cross-referenced equation 3.6.1 with equation 12 from http://mathworld.wolfram.com/CubicFormula.html. I dont think the coefficients are the same. Are they??


    Attached Files:

    Last edited by a moderator: Dec 30, 2014
  2. jcsd
  3. Dec 28, 2014 #2


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    In the complex numbers, z^3 = y always has three distinct solutions (let's ignore the case y=0). Therefore, you get up to 6 solutions. At most 3 are relevant for the final solutions.

    Solutions to the equations discussed above.
    Definition. It is like a statement "if n is an integer, then m=n+1 is an integer": "if v is a solution, then v1=-b/v is a solution" (check this!).

    b and c are not the original b and c, right.
  4. Dec 30, 2014 #3


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    Here is my understanding of Cardano's method: If a and b are any numbers then [itex](a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3[/itex] and [itex]3ab(a+ b)= 3a^2b+ 3ab^2[/itex]. Subtracting, [itex](a+ b)^3- 3ab(a+ b)= a^3+ b^3[/itex].

    Letting x= a+ b, m= 3ab, and [itex]n= a^3+ b^3[/itex] then x satisfies the reduced cubic [itex]x^3+ mx= n[/itex] ("reduced" because it has no squared term). Now, what about the other way around? That is, if we know m and n, can we find a and b and so x?

    Yes, we can. From [itex]m= 3ab[/itex], we have [itex]b= \frac{m}{3a}[/itex] so that
    [tex]n= a^3+ b^3= a^3+ \frac{m^3}{3^3a^3}[/tex]
    Multiply by [itex]a^3[/itex]: [itex]na^3= (a^3)^2+ \left(\frac{m}{3}\right)^3[/itex] or [itex](a^3)^2- na^3+ \left(\frac{m}{3}\right)^3= 0[/itex].

    We can think of that as a quadratic equation in [itex]a^3[/itex] and, by the quadratic formula
    [tex]a^3= \frac{n\pm\sqrt{n^2- 4\left(\frac{m}{3}\right)^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^2}[/tex]
    From [tex]a^3+ b^3= n[/tex]
    [tex]b^3= n- a^3= n- \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^2}=\frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^2} [/tex]

    Take cube roots to find a and b and then x= a+ b. There are, of course, three cube roots for each so 9 possible combinations but some add to give the same x.
  5. Jan 8, 2015 #4


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    Halls' nice explanation is similar to that in the classic text of Euler, pages 262-271, where you may also find several worked examples and some problems. On page 272 ff. he also explains the more complicated case of a quartic.

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