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This is for the page #1,page #2, page #3, page #4 of the book by Dickson, titled "Introduction to Theory of Algebraic Equations", in which there is derivation of solution for general cubic equation.
My question is about not being able to get the desired equality (named $D$, later below) stated at the bottom of page #3.
I request geometrical significance of different root pairs $y_1, y_2, y_3$, so as to understand the way to pursue further.

At the bottom of page $3$ it is given that ( Taking as first equality ($A$) ):
$$\sqrt[3]{\frac{-1}{2}q+\sqrt{R}}= \frac{1}{3}(y_1+\omega (\omega y_2+y_3))$$ $$\frac{-1}{2}q+\sqrt{R}= \frac{1}{27}(y_1+\omega (\omega y_2+y_3))^3 = \frac{1}{27}(y_1^3 +(\omega\cdot y_2+y_3)^3 + 3\omega y_1^2(\omega\cdot y_2+y_3) + 3\omega^2\cdot y_1(\omega\cdot y_2+y_3)^2)$$ $$=\frac{1}{27}(y_1^3 +(\omega^3\cdot y_2^3 +y_3^3 +3\omega^2\cdot y_2^2y_3+3\omega\cdot y_2y_3^2)+3\omega^2 y_1^2y_2+3\omega y_1^2y_3+3\omega^2 y_1(\omega^2\cdot y_2^2+y_3^2+2\omega\cdot y_2y_3))$$ $$=\frac{1}{27}(y_1^3 +y_2^3 +y_3^3+6y_1y_2y_3+3\omega^2\cdot y_2^2y_3+3\omega\cdot y_2y_3^2+3\omega^2 y_1^2y_2+ 3\omega y_1^2y_3 + 3\omega y_1y_2^2+3\omega^2 y_1y_3^2) $$

At the bottom of page $3$ it is next given that ( Taking as second equality ($B$) ):

$$\sqrt[3]{\frac{-1}{2}q-\sqrt{R}}= \frac{1}{3}(y_1+\omega(\omega y_3+y_2))$$ $$\frac{-1}{2}q-\sqrt{R}= \frac{1}{27}(y_1+\omega(\omega y_3+y_2))^3 = \frac{1}{27}(y_1^3 +(\omega\cdot y_3+y_2)^3 + 3\omega y_1^2(\omega\cdot y_3+y_2) + 3\omega^2\cdot y_1(\omega\cdot y_3+y_2)^2)$$ $$=\frac{1}{27}(y_1^3 +(\omega^3\cdot y_3^3 +y_2^3 +3\omega^2\cdot y_3^2y_2+3\omega\cdot y_3y_2^2)+3\omega^2y_1^2y_3+3\omega y_1^2y_2+3\omega^2 y_1(\omega^2\cdot y_3^2+y_2^2+2\omega\cdot y_3y_2))$$ $$=\frac{1}{27}(y_1^3 +y_3^3 +y_2^3+6y_1y_2y_3+3w^2\cdot y_3^2y_2+3\omega\cdot y_3y_2^2+3\omega^2 y_1^2y_3+ 3\omega y_1^2y_2 + 3\omega y_1y_3^2+3\omega^2 y_1y_2^2)$$

At the bottom of page $3$, further the difference of the above equalities is taken up ( Taking the difference as third equality ($C$) ):
$$2\sqrt{R}=\frac{1}{27}((3\omega^2\cdot y_2^2y_3+3\omega\cdot y_2y_3^2+3\omega^2 y_1^2y_2+ 3\omega y_1^2y_3 + 3\omega y_1y_2^2+3\omega^2 y_1y_3^2) -(3\omega^2\cdot y_3^2y_2+3\omega\cdot y_3y_2^2+3\omega^2 y_1^2y_3+ 3\omega y_1^2y_2 + 3\omega y_1y_3^2+3\omega^2 y_1y_2^2))$$

$$\sqrt{R}=\frac{1}{54}(3\omega^2\cdot y_2y_3(y_2-y_3)+3\omega\cdot y_2y_3(y_3-y_2)+3\omega^2 y_1y_2(y_1-y_2)+ 3\omega y_1y_3(y_1-y_3) + 3\omega y_1y_2(y_2-y_1)+3\omega^2 y_1y_3(y_3-y_1))$$

$$=\frac{1}{54}(3\omega(\omega -1)y_1y_2(y_1-y_2)+3\omega (\omega -1)y_2y_3(y_2-y_3)+3\omega(\omega -1)y_1y_3(y_3-y_1))$$

$$=\frac{1}{54}(3(\omega^2-\omega)(y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1)))$$

$$=\frac{(\omega^2-\omega)}{18}(y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1))$$

$$=\frac{\sqrt{-3}}{18}(y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1))$$

How to show the last equality ($C$) equal to the given one below one (let equality $D$) is not apparent to me
$$\sqrt{R}= \frac{\sqrt{-3}}{18}(y_1-y_2)(y_2-y_3)(y_3-y_1)$$

All I could attempt is to take the values of $y_1, y_2,y_3$ and find the products : $y_1y_2, y_2y_3, y_3y_1$. $$y_1= \sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \sqrt[3]{-\frac{q}{2}-\sqrt{R}}$$ $$y_2= \omega \sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \omega^2\sqrt[3]{-\frac{q}{2}-\sqrt{R}}$$ $$y_3= \omega^2 \sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \omega\sqrt[3]{-\frac{q}{2}-\sqrt{R}}$$

$$y_1y_2= (\sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)\cdot (\omega \sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \omega^2\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)$$ $$=\omega \,\,\sqrt[2/3]{-\frac{q}{2}+\sqrt{R}}+\omega^2 \sqrt[2/3]{-\frac{q}{2}-\sqrt{R}}+(\omega+\omega^2)(\sqrt[3]{-\frac{q}{2}+\sqrt{R}}\,\,\,\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)$$

$$y_1y_3= (\sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)\cdot (\omega^2 \sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \omega\,\,\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)$$ $$=\omega^2 \sqrt[2/3]{-\frac{q}{2}+\sqrt{R}}+\omega\,\, \sqrt[2/3]{-\frac{q}{2}-\sqrt{R}}+(\omega+\omega^2)(\sqrt[3]{-\frac{q}{2}+\sqrt{R}}\,\,\,\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)$$

$$y_3y_2= (\omega^2 \sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \omega\,\,\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)\cdot (\omega \sqrt[3]{-\frac{q}{2}+\sqrt{R}} + \omega^2\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)$$ $$=\sqrt[2/3]{-\frac{q}{2}+\sqrt{R}}+\sqrt[2/3]{-\frac{q}{2}-\sqrt{R}}+(\omega+\omega^2)(\sqrt[3]{-\frac{q}{2}+\sqrt{R}}\,\,\,\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)$$ $$=\omega^3 \sqrt[2/3]{-\frac{q}{2}+\sqrt{R}}+ \omega^3 \sqrt[2/3]{-\frac{q}{2}-\sqrt{R}}+(\omega+\omega^2)(\sqrt[3]{-\frac{q}{2}+\sqrt{R}}\,\,\,\sqrt[3]{-\frac{q}{2}-\sqrt{R}}\,\,\,)$$

jiten
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2 Answers2

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Both $y$ expressions expand to $y_1y_2^2-y_1y_3^2-y_2y_1^2+y_2y_3^2+y_3y_1^2-y_3y_2^2$, and $\omega^2-\omega=2i\Im(\omega)=i\sqrt3$.

  • Sorry, could not gather how $w^2 - w = i\sqrt{3}$. – jiten Apr 27 '18 at 10:10
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    @jiten: what is $\omega$ ? –  Apr 27 '18 at 10:12
  • cube root of unity, $\omega ^3 +1=0\implies (\omega+1)(\omega^2 +\omega +1)=0$. $\omega^2 = -(1+\omega)\implies \omega^2 - \omega = - 1- 2\omega$. – jiten Apr 27 '18 at 10:12
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    So what is $\omega^2-\omega$ ? –  Apr 27 '18 at 10:13
  • $\omega$ can be found by the quadratic equation to be: $-\frac{1}{2} \pm \frac{i\sqrt{3}}{2}$ – jiten Apr 27 '18 at 10:19
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    @jiten: don't you get it now ? –  Apr 27 '18 at 10:30
  • Sorry, that was silly. The main issue is how the stated expression as of the differences of equality $A-B=C$, as stated above by: $ y_1y_2^2-y_1y_3^2-y_2y_1^2+y_2y_3^2+y_3y_1^2-y_3y_2^2$ reduces to $(y_1-y_2)(y_2-y_3)(y_3-y_1)$. Please elaborate how to reduce your expression to get equality $D$ (as labeled in the post), & as shown on pg. #3 bottom. – jiten Apr 27 '18 at 10:32
  • I hope my issue is clear, as the reduction of your answer's expression to the needed one is not apparent. – jiten Apr 27 '18 at 10:41
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    @jiten: do you realize that this is elementary algebra ? –  Apr 27 '18 at 10:43
  • $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = (y_1y_2-y_1y_3-y_2^2+y_2y_3)(y_3-y_1)$$ $$=-y_1y_3^2 -y_2^2y_3 +y_2y_3^2-y_1^2y_2+y_1^2y_3+y_1y_2^2$$ $$=y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1)$$ Sorry, have the reversal inside the brackets of terms. All $3$ brackets have opposite terms after the minus sign? – jiten Apr 27 '18 at 11:00
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    Then you've got $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = -[y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1)]$$ – Rhys Hughes Apr 27 '18 at 11:50
  • @RhysHughes But, the book does not seem to use negative sign on page #3 bottom. Also, I don't seem to be wrong in the derivation of equality $C$. If you could prove the sign is wrong in equality $C$, then would be very grateful. – jiten Apr 27 '18 at 11:53
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    @jiten: you should have mentioned a sign error, instead of bombarding with equations. –  Apr 27 '18 at 12:02
  • @YvesDaoust I am not sure what is meant/implied by your last comment. Do you mean that the book is having a sign error, and hence should have harped on that only. – jiten Apr 27 '18 at 12:07
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    @jiten: I mean that you didn't ask a precise question. –  Apr 27 '18 at 12:08
  • @YvesDaoust So, the book is having a sign error? – jiten Apr 27 '18 at 12:08
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There isn't a sign mistake in the book. Note that $\omega - \omega^2 = \sqrt{-3}$.

$$\frac{(\omega^2-\omega)}{18}(y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1))$$

$$=\frac{\color{red}{-}\sqrt{-3}}{18}(y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1))$$

Alternative: \begin{align} &\frac1{54}((y_1+\omega^2y_2+\omega y_3)^3 - (y_1+\omega y_2+\omega^2 y_3)^3)\\ &=\frac1{54}(\omega^2y_2-\omega y_2+\omega y_3 - \omega^2 y_3)((y_1+\omega^2y_2 + \omega y_3)^2+(y_1+\omega^2y_2 + \omega y_3)(y_1+\omega y_2 + \omega^2 y_3)+(y_1+\omega y_2 + \omega^2 y_3)^2)\\ &=\frac{-(\omega - \omega^2)}{54}(y_2-y_3)(3y_1^2+[2(\omega^2y_2 + \omega y_3)+2(\omega y_2 + \omega^2 y_3)+(\omega^2+\omega)(y_2+y_3)]y_1+(\omega^2y_2 + \omega y_3)^2+(\omega^2y_2 + \omega y_3)(\omega y_2 + \omega^2 y_3)+(\omega y_2 + \omega^2 y_3)^2)\\ &= \frac{-\sqrt{-3}}{54}(y_2-y_3)(3y_1^2-3(y_2+y_3)y_1+(\omega y_2^2 + \omega^2 y_3^2+2y_2y_3)+y_2^2+y_3^2+(\omega+\omega^2)y_2y_3+(\omega^2 y_2^2 + \omega y_3^2+2y_2y_3))\\ &=\frac{-\sqrt{-3}}{54}(y_2-y_3)(3y_1^2-3(y_2+y_3)y_1+3y_2y_3)\\ &=\frac{-\sqrt{-3}}{18}(y_2-y_3)(y_1-y_2)(y_1-y_3)\\ &=\frac{\sqrt{-3}}{18}(y_1-y_2)(y_2-y_3)(y_3-y_1)\\ \end{align}

jiten
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Siong Thye Goh
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  • Request significance of this approach getting the first factor as $\omega(\omega-1)(y_2-y_3)$. It seems special that out of three variables, and 3 combinations possible (1.$(y_1,y_2)$, 2.$(y_3,y_2)$, 3.$(y_1,y_3)$) only combination (ii) can be chosen as first factor. --Have a wild guess: multiplying $y_1, y_2, y_3$ by $\omega, 1, \omega^2$ & $\omega^2, 1, \omega$ will get another factor as first; & to get the left combination as first factor, multiply $y_1, y_2, y_3$ by $\omega, \omega^2, 1$ & $\omega^2, \omega, 1$. I hope if correct, would point to geometrical significance of $y_1,y_2,y_3$. – jiten Apr 28 '18 at 09:26
  • Thank a lot for showing the factor theorem at work. Really, all I knew was to multiply the 3 factors $(y_2-y_1)(y_2-y_3)(y_1-y_3)$ to get the expression. Major upgrade for me. – jiten Apr 28 '18 at 09:30
  • $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ – Siong Thye Goh Apr 28 '18 at 15:39
  • I hoped some response for the geometrical significance of the three roots $y_1, y_2, y_3$. May be you want me to place a new post for that. – jiten Apr 28 '18 at 20:00
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    the transformation is a horizontal shift of the original graph such that the new roots sum to $0$. – Siong Thye Goh Apr 28 '18 at 20:08
  • Sorry, could not understand the summing to $0$ part. All I know that in complex cube roots of unity $1+ \omega + \omega^2 = 0$, and these are multiplied to $y_1, y_2, y_3$. The significance of the context of trasformation is also not clear. Are we transforming the 3 roots $y_,y_2,y_3$, if yes then why? Just putting the 3 roots across the 3 cube roots of unity. So, if a quartic equation, then would transform along the four roots of unity? – jiten Apr 28 '18 at 20:17
  • $x=y+\frac{c_1}3$, it is a translation of the graph. we know the sum of $x_i=c_1$, hence the sum of $y_i$ would give us $0$. I am not an expert in this, beyond my ability to help you. possibly related – Siong Thye Goh Apr 28 '18 at 20:21
  • I hoped you meant the sum of all $3 x_i = c_1$. If so, then why the horizontal shift is needed, and what purpose would it serve as per the original intention of Lagrange (the book shows cubic roots solution by him). Also, why the horizontal shift breaks into $3$ equal parts. Then how this horizontal shift is possible to be mapped into the $3$ roots of unity. – jiten Apr 28 '18 at 20:24
  • I am not the right person to consult. – Siong Thye Goh Apr 28 '18 at 20:27
  • Okay, then at least that was a typo... i.e. you meant $3x_i = c_1$. – jiten Apr 28 '18 at 20:29
  • Let us continue this discussion in chat. – Siong Thye Goh Apr 28 '18 at 20:30
  • I edited your answer again today, but was rejected. I request if there was a mistake in it. Its image is at :https://i.stack.imgur.com/nQLSj.png. – jiten May 01 '18 at 10:28
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    not really but usually edit has to go through review and it is more likely to be approved if it is just minor typographical edit. so just expansion of $y$'s right? and yup, it doesn't seem like you are using factor theorem in your original working. – Siong Thye Goh May 01 '18 at 15:20
  • I have an unusual request for you, need to chat for guidance. I hope you would understand my urgency. – jiten May 11 '18 at 18:12
  • hmm... ok, which chat room? – Siong Thye Goh May 11 '18 at 18:27
  • older one is oay, i.e. https://chat.stackexchange.com/rooms/77291/discussion-between-siong-thye-goh-and-jiten – jiten May 11 '18 at 18:28
  • Have sent a mail to you from [email protected]. Please check. – jiten May 14 '18 at 15:24
  • Please check my post at : https://math.stackexchange.com/q/2785844/424260. It concerns with uniform model used to generate rectangles using the point model, where only one point need be stated, with the other being origin. – jiten May 18 '18 at 03:59
  • Please help with my earlier comment's post. No response till now from any. Stuck. – jiten May 18 '18 at 05:47