how did Cardano obtain three solutions for cubic?

Question

So, if I am not mistaken Complex numbers were discovered after Cardano's method. But from Cardano's Method on Wikipedia, it says to get the three solutions, we should use the root of unity. In that case, when Cardano's method was published it could only find one solution?

Answer 1

For example, to find the cube roots of unity, take the solutions of the equation $$x^3=1,$$ which can be found by writing $$x^3-1=0\longrightarrow(x-1)(x^2+x+1)=0.$$

Cardano's method produces a similar result for a certain class of cubic equations.

The root of $x-1$ of course is real but the others are nonreal conjugates.

Answer 2

I am interpreting the question to be historical: how could Cardano's method give the complex solutions before complex numbers had been invented? Here is the answer:

Cardano's method as described in Cardano's 1545 book Ars Magna is only for finding a (single) real solution.

Ironically, this book also contains the first published account of complex numbers; however, not in the same part of the book that describes the solution to cubics. (Actually the solution to the cubics are the bulk of the book.) Cardano does not apply complex numbers to cubics in the book.

27 years later, Rafael Bombelli published a book that explicitly tied the idea of imaginary numbers to the solution of the cubic equation. My understanding however is that his application was to show that introducing $i$ allows one to use Cardano's formula even in the casus irreducibilis and not to use roots of unity to show that Cardano's formula really gives three solutions.

When the Wikipedia article describes Cardano's solution, it is a modern interpretation of Cardano's actual work. The three solutions that it gives come from this modern interpretation. Mathematicians didn't really get comfortable with complex numbers until the 18th and 19th centuries.

Answer 3

The method of solving a cubic is explained very clearly in Stroud's Further Engineering Mathematics. According to Stuart Hollingdale's book "Makers of Mathematics", Cardano plagiarized Tartaglia's solution of the cubic. Paul Halmos said that he solved this problem for himself as a high school student.

Answer 4

From $x^3+px+q=0$ writing $x=u+v$ gives

(1) $$u^3+v^3=a$$ (2) $$3uv=b\Rightarrow u=\frac{b}{3v}$$

Plugging in (1) gives a bicubic :

(3) $$\frac{b^3}{27}+v^6=av^3$$

Leading to 2 values of $v^3$

To find $u^3$ from $v^3$ there are 2 ways (1) gives one possibility for $u^3$, the sum $u+v$ has hence 3x3 possibilities for the cube roots, the 2 values for (3) don't matter here since it has a form $\sqrt[3]{w\pm z}+\sqrt[3]{w\mp z}$

The second way leads to 3x3 possibilities but here the 2 cases of (3) matter.

Hence $x=u+v$ can take 27 possible values. Each of them has to be checked again since only 3 are possible.

For example $x^3-x-1=0$

Cardano's method is :

Make $x=u+v$, hence

$$u^3+v^3-1+(u+v)(3uv-1)=0$$

System obtained (1) $u^3+v^3=1$ and (2) $u=1/(3v)$

Substituting 2 in 1 gives

$$1/27+v^6-v^3=0$$

(3) $$v^3=(1\pm\sqrt{1-4/27})/2$$

Now here comes the 2 ways :

1. $u^3=1-v^3=(1\mp\sqrt{1-4/27})/2$ Here $u+v$ gives one possibility (outside complex phases)

2. $u=1/(3v)$ leads to 2 possibilities when switching the sign for the value of $v$ obtained in (3).