Understanding Cardano's method of solving Cubic equation

[PcumP_Ravenclaw]

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 don't think the coefficients are the same. Are they??

Danke...

 

[mfb]

z3=y1ory2 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 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.

what is v and v1.

Solutions to the equations discussed above.

why is v1 = -b/v

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.

 

[HallsofIvy]

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.

 

[mathwonk]

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.

https://archive.org/details/elementsofalgebr00eule

 

[blue_raver22]

Hello,

Thank you for reaching out and expressing your interest in understanding Cardano's method for solving cubic equations. I will try my best to explain the concepts and answer your questions.

Firstly, it is important to note that Cardano's method is one of the methods used to solve cubic equations. It is named after the Italian mathematician Gerolamo Cardano who first published it in his book "Ars Magna" in 1545. The method involves making a series of substitutions and manipulations to transform the original cubic equation into a simpler form that can be solved using known methods.

Now, let's address your questions one by one.

1. The two substitutions you mentioned, ## P1(z - a/3)## and ##P(z - b/z) ##, are indeed part of the process in Cardano's method. The first substitution is used to eliminate the quadratic term in the original cubic equation (##z^3 + az^2 + bz + c = 0##) and the second substitution is used to eliminate the cubic term. This results in the equation you mentioned, ##z^6 - cz^3 - b^3 = 0##, which can then be simplified further.

2. You are correct in saying that the solutions for ##z## in the equation ## z^3 = y1## or ##y2 ## are only two, namely 2 and -2. However, when we substitute these values back into the original cubic equation, we get three distinct solutions for ##z##, as you can see in the paragraph from the text. This is because the other solutions for ##z## (from ##z^6 - cz^3 - b^3 = 0##) are complex numbers, and when we take their cube root, we get three distinct solutions.

3. The variables ##v## and ##v1## are introduced in the text to represent the roots of the equation ##z^6 - cz^3 - b^3 = 0##. This is a result of the substitution ##P(z - b/z) ##, which leads to a quadratic equation in terms of ##z^3##. The roots of this quadratic equation are ##v## and ##v1##, and as you can see, they are related by ##v1 = -b/v##.

4. The coefficients, a, b, and c, in the equation 3.6.1 of