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This is in continuance to my earlier post : Deriving expression for general cubic equation solution. Also, the links are repeated for the first four pages of the book by Dickson, titled "Introduction to Theory of Algebraic Equations" are given as : page #1,$\,$page #2,$\,$ page #3, $\,$page #4.$\,$ Added are rest $5$ pages (#$5$ to #$9$) from first chapter of the book : page #5, $\,$ page #6,$\,$ page #7,$\,$ page #8,$\,$ page #9.

On page $4$, the book states that: "Aside from the factor 1/3, the roots of the sextic ($5$) (i.e, $z^6 +qz^3 -\frac{p^3}{27}=0$) are $\cdots$"
But, how the value 1/3 is a root of the sextic (5) is not clear.

I will be putting stated factor $\frac{1}{3}$ in equation $(5)$ to get:
$(\frac{1}{3})^6 +q(\frac{1}{3})^3 -\frac{p^3}{27}=0 => \frac{1}{27^2} +q(\frac{1}{27})- \frac{p^3}{27} = 0$

The values of $p =c_2 - \frac{1}{3} c_1^2 , q = -c_3+\frac{1}{3}c_1c_2 -\frac{2}{27}c_1^3$ are of constants and are dependent only on the coefficients of the general cubic equation (stated on page $1$ as : $x^3 -c_1x^2 +c_2x +c_3=0$), hence cannot be simplified further.

jiten
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1 Answers1

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It doesn't mean that $\frac13$ is the root of the cubic equation.

We already know that $$z^3 =-\frac12q\pm\sqrt{R}$$

and $$\sqrt[3]{-\frac12q+\sqrt{R}}=\color{blue}{\frac13}\left(y_1+\omega^2y_2+\omega y_3 \right)=\color{blue}{\frac13}\left(x_1+\omega^2x_2+\omega x_3 \right)$$

What it meant is that $\frac13 \psi_i$ are the roots of $$z^6+qz^3-\frac{p^3}{27}=0.$$

Siong Thye Goh
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  • Thanks, and would request you to state any reason for Hudde substitution (on page #$2$ of the book, $y = z - \frac{p}{3z}$) that specifically uses $p$, that I hope stands for $p=c_2 -\frac{1}{3}c_1^2\implies y= z-\frac{(x_1x_2+x_2x_3+x_3x_1)-\frac{1}{3}(x_1+x_2+x_3)^2}{3z}$$\implies y = z -\frac{x_1(x_2-x_1)+x_3(x_1-x_3)+x_2(x_3-x_2)}{3z}$. In my all questions so far, I have not been aware of this issue, and the book sidelines it too. – jiten May 03 '18 at 07:58
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    my current understanding of the approach is superficial. I am waiting for a cool answer as well. I can't see it. – Siong Thye Goh May 03 '18 at 08:01
  • Kindly find a new post on this issue at : https://math.stackexchange.com/q/2764553/424260. Kindly suggest edits, if needed. – jiten May 03 '18 at 08:10
  • Have found some simple reference to Hudde's substitution in a book titled : Patterns of Change: Linguistic Innovations in the Development of Classical ... By Ladislav Kvaszin, in $4$ pages, may be it helps somewhat. The $4$ pages of concern are (i) page 177-178: https://i.stack.imgur.com/SJaAX.png; (ii) page 179-180: https://i.stack.imgur.com/if9sD.png It comes quite near in pointing out 'clearly' on page #$178$ in the quadratic form, the product of the two factors, i.e. $c_2$ is same as the cube root of last term ($\frac{-p}{3}$). – jiten May 03 '18 at 08:56
  • Want to clarify that apart from a general explanation of Hudde's substitution, the last comment, has 'no' concern with the latest problem as stated earlier. Also, I hope that a new page ($366$) from the book : Algebra in the mirrors of mathematics, by Gabriel Katz may help in some improvement in understanding of the workings of the cardano's formula for cubic: https://i.stack.imgur.com/2LPcT.png. In fact, it seems that Hudde made a guess to make things work, and the logic (as in my first comment) can be shown only by curve drawing. – jiten May 03 '18 at 17:43
  • Sorry, not worthy of separate question. Want plot points $M,H,C,R,A,B$ in paper: http://i.stack.imgur.com/EVRu1.png, in desmos: https://www.desmos.com/calculator/m7kgvinue1, with analysis: $(i)$ Need two points for some secant line (that will help form parametric tangent/secant line for curve) with $y'=0$ for $x_i=\pm 2$, (ii) $y_i= -6,-2$, with slope of secant between first ($H$) & second cusp (let $Z$) with slope $=-1$ & eqn. of secant as $y=-x+c$. That did not work. Seems need parametric curve with two initial points:$(i) R, y=0$,(ii)$H$. So I need Cardano's formula to get $x$ for $R$? – jiten May 09 '18 at 02:27
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    depending on your purpose, there isn't a need to do everything exactly, as we can read off the coordinate of $R$ from your current plot. – Siong Thye Goh May 09 '18 at 03:01
  • I am sorry for not stating the problem correctly. I want to plot a parametric curve that will be able to draw secant, tangent depending on some parameter. – jiten May 09 '18 at 03:14
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    sure, you can use Cardano. – Siong Thye Goh May 09 '18 at 03:32
  • Please see my posts at: (1) https://math.stackexchange.com/q/2804677/424260. I am confused with reason shown in my last comment to the OP, and to the highest voted answer.(2) https://math.stackexchange.com/q/2805021/424260. This is about using the Viete formula to solve the equation. – jiten Jun 02 '18 at 04:01