Why is 3AB+P Set to Zero in Cardan's Cubic Equation?

[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?

 

[tiny-tim]

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

$$X_0 = A + B$$

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

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

The next step is 3AB+P=0

What make Cardano drop 3AB+P?

Why it is zero?

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).

And we already know how to solve a quadratic!

 

[Physicsissuef]

Ok, thank you very much Timmy.

 

[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