The Expressions for Roots of Cubic Polynomial // تعبيرات جذور كثيرة الحدود التكعيبية، مرة أخرى بمساعدة اللاهوت

                We have discussed this issue before. We will discuss the similarities and differences with the current situation at the end.

          Here we consider the problem : 

                   Let  $\tau$  be one of the roots of the equation

$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{1}$

                   The other two roots are

$\tau_2=\frac{\sqrt{3}}{2}\cdot \tau^2-5\cdot \tau+\frac{\sqrt{3}}{2}\;,\;\;\tau_3=-\frac{\sqrt{3}}{2}\cdot \tau^2+4\cdot \tau+\frac{5\sqrt{3}}{2} \tag{2}$ 

SOLUTION  CiP

               From the fact that  $\tau$  checks  (1)  we obtain some useful formulas in future calculations

$\tau^3=3\sqrt{3}\cdot \tau^2+3\cdot \tau -\sqrt{3} \tag{3}$

$\tau^4=30\cdot \tau^2+8\sqrt{3} \cdot \tau-9 \tag{4}$

$\frac{1}{\tau}=-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3} \tag{5}$

          Indeed, $\tau^3-3\sqrt{3}\cdot \tau^2-3\cdot \tau +\sqrt{3}=0 \Rightarrow \tau^3=3\sqrt{3} \cdot \tau^2+3\cdot \tau-\sqrt{3}$,  i.e. (3). Then  $\tau^4=\tau \cdot \tau^3\overset{(3)}{=}$

$=\tau (3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})=3\sqrt{3}\cdot \tau^3+3\cdot \tau^2-\sqrt{3}\cdot \tau\overset{(3)}{=}3\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+3\cdot \tau^2-\sqrt{3}\cdot \tau=30\cdot \tau^2+8\sqrt{3}\cdot \tau-9$ i.e. (4). Finally, $\sqrt{3}=-\tau^3+3\sqrt{3}\cdot \tau^2+3\cdot \tau=\tau \cdot (-\tau^2+3\sqrt{3}\cdot \tau+3)\Rightarrow\frac{1}{\tau}=\frac{-\tau^2+3\sqrt{3}\cdot \tau+3}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot \tau^2+3\cdot \tau+\sqrt{3}$, i.e. (5).

          From the first formula of Vieta  $\tau_1+\tau_2+\tau_3=3\sqrt{3}$  we get

$$\tau_2+\tau_3=3\sqrt{3}-\tau \tag{6}$$

and from the third  $\tau_1\cdot \tau_2\cdot \tau_3=-\sqrt{3}$  we deduce

$$\tau_2\cdot \tau_3=-\frac{\sqrt{3}}{\tau} \tag{7}$$

(6) and (7) show that  $\tau_{2,3}$  are the roots of the quadratic equation

$t^2-(3\sqrt{3}-\tau) \cdot t-\frac{\sqrt{3}}{\tau}=0 \tag{8}$

          Let's calculate its discriminant:  $\Delta_2=(3\sqrt{3}-\tau)^2+\frac{4\sqrt{3}}{\tau}=$

$\overset{\color{Red}{!!!}}{=}27-6\sqrt{3}\cdot \tau+\tau^2+\frac{4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{\tau^4-6\sqrt{3}\cdot \tau^3+27\cdot \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=$

$\overset{(4)}{\underset{(3)}{=}}\frac{(30\cdot \tau^2+8\sqrt{3}\cdot \tau-9)-6\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+27 \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{3\cdot \tau^2-6\sqrt{3}\cdot \tau+9}{\color{Red}{\tau^2}}=\frac{(\sqrt{3}\cdot \tau-3)^{\color{Red}2}}{\color{Red}{\tau^2}}$

<  Hallelujah, I got a perfect square  !!!  > 

Let's do a little more calculation to make writing easier :  $\pm\sqrt{\Delta_2}=\frac{\sqrt{3}\cdot \tau-3}{\tau}=$

$=(\sqrt{3}\cdot \tau-3)\cdot \frac{1}{\tau}\overset{(5)}{=}(\sqrt{3}\cdot \tau-3)\left (-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3}\right )=$

$=-\tau^3+\sqrt{3}\cdot \tau^2+3\sqrt{3}\cdot \tau^2-9\cdot \tau+3\cdot \tau-3\sqrt{3}=$

$\overset{(3)}{=}-(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+4\sqrt{3}\cdot \tau^2-6\cdot \tau-3\sqrt{3}=\color{Blue}{\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3}}$

 !!! Remember this expression !!!

With these, the equation  (8)  has the roots

$\tau_{2,3}=\frac{3\sqrt{3}-\tau \pm \sqrt{\Delta_2}}{2}=\frac{3\sqrt{3}-\tau \pm (\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})}{2} \tag{9}$

from where we obtain the expressions  (2).

$\blacksquare$

               REMARKS CiP

                    1.  In the calculation of  $\Delta_2$  that follows immediately after the equation  (8), I made a trick so that at least the denominator would be a perfect square. Then I was lucky enough !!  Compare with the Post  << Not only God does not help, but also Allah does not "put you in trouble">>.

If in (7) we had used (5), obtaining  $\tau_2\cdot \tau_3=\tau^2-3\sqrt{3}\cdot \tau-3$, then for the equation (8) we would have had  $\Delta_2=(3\sqrt{3}-\tau)^2-4(\tau^2-3\sqrt{3}\cdot \tau-3)=39+6\sqrt{3}\cdot \tau -3\cdot \tau^2.$  Seeing this expression of  $\Delta_2$  we wondered, as we did in the Post How? can we recognize a perfect square??, "how the hell" can we notice - compare with (9) - that

$$\color{Blue}{39+6\sqrt{3}\cdot \tau-3\cdot \tau^2=(\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})^2}$$

Problem NOT solved yet.

                    2.  The theoretical part about when and how such a statement is possible was exposed in  the  Post "The Rational Expressions for Roots of Cubic Polynomial". There are some differences, but only apparent.

          First of all, that the polynomial  $f=X^3-3\sqrt{3}\cdot X^2-3\cdot X+\sqrt{3}$  is not in  $\mathbb{Q}[X]$  but in  $\mathbb{Q}(\sqrt{3})[X].$

          Then we need to make sure that the polynomial $f$  is irreducible. Maybe this is not that important, and I don't have an immediate justification at hand. 

          Third, its discriminant must be a perfect square (see Proposition in the Post). From  $f$  we obtain the depressed cubic by substitution  $t=u+\sqrt{3}$, obtainig the polynomial  $u^3-12\cdot u-8\sqrt{3}$. Its discriminant, the same as for  $f$, is  $-4p^3-27q^2=4\cdot12^3-27\cdot(64\cdot3)=1\;728$  so equal to $3\cdot 576=(24\sqrt{3})^2$, hence perfect square in  $\mathbb{Q}(\sqrt{3})$.

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