how did Cardano obtain three solutions for cubic? 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?  

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 :

*

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


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