Cardan's cubic equation

1. Sep 6, 2008

Physicsissuef

$$X^3+PX+Q=0$$

$$X_0 = A + B$$

$$(A+B)^3 + P(A+B) + Q = 0$$

$$A^3 + B^3 + (3AB+P)(A+B) + Q = 0$$

The next step is 3AB+P=0

What make Cardano drop 3AB+P?

Why it is zero?

2. Sep 6, 2008

tiny-tim

Hi Physicsissuef!

AB can be anything we like, so long as A + B = X.

So we choose a ratio such that AB = -P/3.

That eliminates one of the terms in the equation, and makes it much easier … a quadratic equation (in A^3, I think).

3. Sep 7, 2008

Physicsissuef

Ok, thank you very much Timmy.

4. Oct 17, 2010

kennysmith39

could someone just show me with an example on how to work the cardano and Del ferro method from start to finish say using
x^3-6x^2+2x-1