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

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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?
 
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Physicsissuef said:
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! :smile:

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! :wink:
 
Ok, thank you very much Timmy.
 
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
 

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