Understanding the method to solve the cubic equation

[PcumP_Ravenclaw]

Dear All,
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 don't think the coefficients are the same. Are they??

Danke...

 

[mathwonk]

try euler's explanation, i like it better:

Next I want give Euler's explanation of how to solve cubic equations. first he shows that any cubic equation can be transformed by a trick to change the cubic into one with no X^2 term. So it is only necessary to be able to solve cubics like this one: X^3=pX+q. E.g.supposethatwehaveX^3=9X+28.

Then Euler explains that to solve this all we need to do is find two numbers u,vsuchthat3uv=9andu^3+v^3=28. ThenX=u+vwillsolvethe cubic.

14. See if you can use Euler's method to solve X^3 = 9X + 28.

(Essentially this solution method was found apparently by Scipio del Ferro, and later Tartaglia, who explained it to Girolamo Cardano, who eventually published it. It is often called "Cardano's method".)

15. Try this one as well: X^3 = -18X + 19.Solving cubics used to be taught in elementary algebra books, but when I went to high school, it was no longer done, and I think is not commonly done now. This is another reason to prefer the great old algebra books like those of Euler and LaGrange.

16. Show this method can be used on all cubics of form X^3 = pX + q.

I.e. i) show that one can always find numbers u,v such that u^3+v^3 = q, and 3uv = p.

(Hint: let A=u^3 and B=v^3. Then you know that A+B = q, and AB = p^3/27; why? Then show how to find A and B, and then tell how to find u and v.)

ii) ShowthatifX=u+vandif3uv=pandu^3+v^3=q,thenX^3=pX +q.

The whole point of understanding quadratic equations is this. Write them like X^2-bX+c = 0, and then b is always the sum of the solutions and c is always their product.

Vice versa, whenever you are looking for two numbers and you already know their sum and their product, then you can always find the numbers as the solutions of a quadratic equation.

The whole point of solving cubics is this: First we know from studying quadratics that we can always find two numbers whose sum and product are known,. Next Euler shows that to solve the cubic X^3 = pX + q, all you need is two numbers u,v such that u^3 +v^3 = q, and 3uv = p.

So we just need to find u and v. But it turns out we already know how to find their cubes, u^3 and v^3.

I.e. what do we know about the cubes of u,v? we know their sum u^3 + v^3 = q, and we know their product, since 3uv = p, so 27 u^3v^3 =p^3, hence u^3v^3 = p^3/27.

So since we know the sum and product of u^3 and v^3, we can find u^3 and v^3 by solving a quadratic! Then we can take cube roots to find u and v.

I.e. if r is a root of the quadratic t^2 - qt + (p^3/27), set u = cuberoot(r) and v = p/3u, and then u + v = X solves the cubic!