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PcumP_Ravenclaw

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Dear All,

I am trying to understanding Cardano's method to solve the cubic equation. Please first see the reference text that 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...

I am trying to understanding Cardano's method to solve the cubic equation. Please first see the reference text that 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...

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