In the first major studies into the magical properties of the complex number and the application to solving the cubic equation, Cardano did the following in around 1539:

He reduced the cubic equation


to the form


where $p$ and $q$ are real numbers.

(1). How did Cardano get to this?

There was also another solution to the cubic, discovered before 1926, and it is often referred to as the (del Ferro-)Cardano solution, and is perhaps where Cardano went from in his first reduction. This solution is:


where $w=\sqrt{(q^2-p^3)}$.

(2). How is the del Ferro-Cardano solution derived?


3 Answers 3


Deriving cubic formula using Cardano's Method :

The general cubic equation is $x^3+bx^2+cx+d=0$ We set $x = y -\frac{b}{3}$ ( This is done to get rid of the component $bx^2$) with the help of this we get the reduced cubic equation in the form :

$y^3+py +q=0.......(i) $ ( Find p and q )

Now we introduced two variables u and v linked by the condition $ u +v = y$ Substitute this in the depressesd cubic , we get :

$u^3+v^3+ (3uv+p)(u+v) +q =0$.......(ii)

Now if $3uv+p =0$ then (ii) becomes : $u^3+v^3 =-q$ and $u^3v^3 = \frac{-p^3}{27}$

We can say that $u^3 ; v^3$ are two roots of equation $z^2+qz -\frac{p^3}{27}=0$

Hope from here you can get your result.... in case further clarification required please let me know.


1) To go from the general cubic equation to the normal form one performs the transformation $x = \alpha t+\beta$. A good choice of $\alpha$ and $\beta$ yields $t^3 = 3px + 2q$. One chooses $\alpha$ and $\beta$ so that the other terms drop out.

2) I believe the solution to your problem is given on wikipedia.


As to part (2), compare $$ x^3 = 3 p x + 2 q $$ with the expression $$ (u + v)^{3} = 3 u v (u + v) + u^3 + v^3. $$ One is led to try to set $x = u + v$, and try and solve $uv = p$ and $u^3 + v^3 = 2 q$, or $$ u^3 + v^3 = 2 q, \qquad u^3 v^3 = p^3. $$ Thus $u^3, v^3$ are roots of $$ z^2 - 2 q z + p^3, $$ and this has solutions $$ z = q \pm \sqrt{q^2 - p^3}, $$ hence Cardano's solution.


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