Dear All,(adsbygoogle = window.adsbygoogle || []).push({});

I am trying to understanding Cardano's method to solve the cubic equation. The reference text is attached (Alan F. Beardon, Algebra and Geometry).

My rough understanding is that we make 2 substitutions (## P1(z - a/3) and P(z - b/z) ## ) to simplify the cubic equation to the from (## z^6 -cz^3 - b^3 = 0 ##) so that we can make a further substitution (## y = z^3##). This results in a quadratic equation which can be solved to get two solutions y1 and y2.

## z^3 = y1 or y2 ## say y1 and y2 are 8 and -8 then this results in only two solutions right 2 and -2. Without forgetting that we need to do 2 Reverse substitutions to get the final solution. Why does the book say 6 solutions for z and only 3 are distinct?

In the paragraph below from the text.

" However, the values of ζ are the

roots of the equation ## z^6 − cz^3 − b^3 = 0## , and if v is a root of this equation,

then so (trivially) is v1 = −b/v. As v1 − b/v1 = v − b/v, we see that these six

values of ζ can only provide at most three distinct roots of p. "

what is v and v1. why is v1 = -b/v (The substitution was ##P(z - b/z) ## ) ??

I think the coefficients, a, b & c from the equation 3.6.1 and the equation below 3.6.1 in the text are NOT the same as the original cubic equation (## z^3 + az^2 + bz + c = 0 ##). I cross-referenced equation 3.6.1 with equation 12 from http://mathworld.wolfram.com/CubicFormula.html. I dont think the coefficients are the same. Are they??

Danke...

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Understanding the method to solve the cubic equation

**Physics Forums | Science Articles, Homework Help, Discussion**