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Master Lagrange (1736 – 1813) revealed at least two centuries ago primitive third roots of unity in terms of the exponents as the solutions of the equation:

$\zeta ^{3}-1=0=(\zeta -1)(\zeta ^{2}+\zeta +1)=(\zeta -1)\left(\zeta -{\frac {i{\sqrt {3}}-1}{2}}\right)\left(\zeta -{\frac {2}{i{\sqrt {3}}-1}}\right){\text{ yielding}}$ ${\sqrt[{3}]{1}}=\zeta ^{0;\pm 1}=\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)^{0;\pm 1}=1;-{\frac {1}{2}}\pm i{\frac {\sqrt {3}}{2}}{\text{ or in trigonometric notation }}{\sqrt[{3}]{1}}=\cos {\frac {2k\pi }{3}}+i\sin {\frac {2k\pi }{3}}=e^{\frac {2ik\pi }{3}}.$

The formula in terms of the coefficients can be obtained by means of the section 3.5 Vieta’s (1540 - 1603) substitution which is properly sourced at the works of Nickalls R. W. D. (see Notes 16.” Viète, Descartes and the cubic equation” and 29. "A new approach to solving the cubic”).

Therefore I see no need essentially the same approach to be repeated hence the sections 2 (along with the image NR. 3), 3.2, 3.3, 3.5, 3.7 and 3.8 can be rearranged and titled at following order:

2 Derivatives, the function flow and reduction to depressed cubic equation

The notation implemented here is geometrically grounded hence the variables are the abscise (x_S), ordinate (y_S = aq) and slope (m_S = ap) at the point of the inflection and Symmetry S(x_S,y_S) where: $x_{S}={\frac {-b}{3a}},y_{S}=f(x_{S})=ax_{S}^{3}+bx_{S}^{2}+cx_{S}+d={\frac {2b^{3}-9abc+27a^{2}d}{27a^{2}}}{\text{ and }}m_{S}=f'(x_{S})=3ax_{S}^{2}+2bx_{S}+c={\frac {3ac-b^{2}}{3a}}.$ ${\text{Inserting }}f^{(3)}(x_{S})=6a,f''(x_{S})=6ax_{S}+2b=0,f'(x_{S})=m_{S}{\text{ and }}f(x_{S})=y_{S}{\text{ at Taylor’s series about S we get}}$ $f(x)={\frac {6a}{3!}}(x-x_{S})^{3}+{\frac {m_{S}}{1!}}(x-x_{S})+{\frac {y_{S}}{0!}}=a(x-x_{S})^{3}+m_{S}(x-x_{S})+y_{S}{\text{ being depressed cubic function in }}x-x_{S}={\frac {3ax+b}{3a}}.$ ${\text{ Last two items form }}f_{t}(x)=\tan \sigma (x-x_{S})+y_{S}{\text{ which is a tangent of }}f(x){\text{ at point S where }}\tan \sigma =f'(x_{S})=m_{S}.$ ${\text{See graph of }}f(x)=x^{3}-3x^{2}-144x+432=(x-1)^{3}-147(x-1)+286{\text{ and }}f_{t}(x)=-147x+433=-147(x-1)+286.$

${\text{Besides, }}f'(x)=3a(x-x_{S})^{2}+m_{S}=3(x-1)^{2}-147=0{\text{ determines }}x_{max;min}=x_{S}\mp {\sqrt {\frac {m_{S}}{-3a}}}=1\mp 7{\text{ existing if }}{\frac {m_{S}}{a}}<0.$

3 Vietè’s substitution and real solution in terms of the inflection point properties

$a(x_{0}-x_{S})^{3}+m_{S}(x_{0}-x_{S})+y_{S}=0{\text{ or }}(3ax_{0}+b)^{3}-3(b^{2}-3ac)(3ax_{0}+b)=9abc-2b^{3}-27a^{2}d$

is depressed cubic equation for which we can find real solution (x₀) by means of

$x_{0}-x_{S}=z-{\frac {m_{S}}{3az}}{\text{ being Vieta's substitution for this form of the equation yielding quadratic one in }}z^{3}$ $(z^{3})^{2}+{\frac {y_{S}}{a}}z^{3}-\left({\frac {m_{S}}{3a}}\right)^{3}=0{\text{ satisfied for }}z^{3}={\frac {-y_{S}}{2a}}+{\sqrt {\left({\frac {-y_{S}}{2a}}\right)^{2}+\left({\frac {m_{S}}{3a}}\right)^{3}}}{\text{ that after rooting and back-conversion gives}}$ $z={\sqrt[{3}]{{\frac {-y_{S}}{2a}}+{\sqrt {\left({\frac {-y_{S}}{2a}}\right)^{2}+\left({\frac {m_{S}}{3a}}\right)^{3}}}}}{\text{ and }}x_{0}=x_{S}+{\sqrt[{3}]{{\frac {-y_{S}}{2a}}+{\sqrt {\left({\frac {-y_{S}}{2a}}\right)^{2}+\left({\frac {m_{S}}{3a}}\right)^{3}}}}}+{\sqrt[{3}]{{\frac {-y_{S}}{2a}}-{\sqrt {\left({\frac {-y_{S}}{2a}}\right)^{2}+\left({\frac {m_{S}}{3a}}\right)^{3}}}}}$ being real number even if the item under square root is negative as confirmed at section 3.2 below implementing polar coordinate system.

Now is proper moment for introducing any of above quoted expression for primitive third roots of unity where we can choose any set of three consecutive integers –1; 0; +1 or 999; 1000; 1001 but I intercede for 0; 1; 2 in order a coherency with a majority of the sections (3.6, 3.7 and 3.9.1) to be maintained. If so, t₁ in section 3.4 Cardano’s method should be replaced by t₀ enabling x₀ to be real solution within entire article.

3.1 Factorization enabling common formula in terms of the inflection point properties

Annuling the quotient of the differences of ordinates and abscises we get

${\frac {a(x-x_{S})^{3}+m_{S}(x-x_{S})+y_{S}-a(x_{0}-x_{S})^{3}-m_{S}(x_{0}-x_{S})-y_{S}}{a(x-x_{S})-a(x_{0}-x_{S})}}=(x-x_{S})^{2}+(x_{0}-x_{S})(x-x_{S})+(x_{0}-x_{S})^{2}+{\frac {m_{S}}{a}}=0$

${\text{for }}x_{1:2}=x_{S}-{\frac {1}{2}}(x_{0}-x_{S})\pm i{\frac {\sqrt {3}}{2}}{\sqrt {(x_{0}-x_{S})^{2}+{\frac {4m_{S}}{3a}}}}{\text{ that inserting Vieta’s substitution }}x_{0}-x_{S}=z-{\frac {m_{S}}{3az}}{\text{ becomes}}$

$x_{1:2}=x_{S}-{\frac {1}{2}}\left(z-{\frac {m_{S}}{3az}}\right)\pm i{\frac {\sqrt {3}}{2}}\left(z+{\frac {m_{S}}{3az}}\right)=x_{S}+\left(-{\frac {1}{2}}\pm i{\frac {\sqrt {3}}{2}}\right)z-\left(-{\frac {1}{2}}\mp i{\frac {\sqrt {3}}{2}}\right){\frac {m_{S}}{3az}}=x_{S}+e^{\pm {\frac {2i\pi }{3}}}z-e^{\mp {\frac {2i\pi }{3}}}{\frac {m_{S}}{3az}}.$

Hence z is now known all of these three solutions can be merged into an algebraic formula in terms of the inflection point properties

$x_{k}=x_{S}+e^{\frac {2ik\pi }{3}}{\sqrt[{3}]{{\frac {-y_{S}}{2a}}+{\sqrt {\left({\frac {-y_{S}}{2a}}\right)^{2}+\left({\frac {m_{S}}{3a}}\right)^{3}}}}}+e^{\frac {-2ik\pi }{3}}{\sqrt[{3}]{{\frac {-y_{S}}{2a}}-{\sqrt {\left({\frac {-y_{S}}{2a}}\right)^{2}+\left({\frac {m_{S}}{3a}}\right)^{3}}}}}{\text{ for }}k=0;1;2.$

3.2 Algebraic and trigonometric formula for all of three solutions in terms of the coefficients

In order to shorten the algebraic formula to the width of A4 hard copy after inserting of $x_{S}={\frac {-b}{3a}},y_{S}=f\left({\frac {-b}{3a}}\right){\text{ and }}m_{S}=f'\left({\frac {-b}{3a}}\right)={\frac {3ac-b^{2}}{3a}}{\text{ we, taking out }}{\sqrt[{3}]{\frac {-y_{S}}{2a}}}={\frac {1}{3a}}{\sqrt[{3}]{\frac {9abc-2b^{3}-27a^{2}d}{2}}},{\text{ get}}$ $x_{k}=\left({\frac {e^{\frac {2ik\pi }{3}}}{3a}}{\sqrt[{3}]{1+{\sqrt {1-{\frac {4(b^{2}-3ac)^{3}}{(9abc-2b^{3}-27a^{2}d)^{2}}}}}}}+{\frac {e^{\frac {-2ik\pi }{3}}}{3a}}{\sqrt[{3}]{1-{\sqrt {1-{\frac {4(b^{2}-3ac)^{3}}{(9abc-2b^{3}-27a^{2}d)^{2}}}}}}}\right){\sqrt[{3}]{\frac {9abc-2b^{3}-27a^{2}d}{2}}}-{\frac {b}{3a}}$ ${\text{ which is apparently complex if }}4(b^{2}-3ac)^{3}>(9abc-2b^{3}-27a^{2}d)^{2}>0{\text{ when conversion to polar coordinates should}}$ ${\text{ be applied by means of }}z=e^{i\theta }{\sqrt {b^{2}-3ac}}{\text{ and }}{\frac {9abc-2b^{3}-27a^{2}d}{2{\sqrt {(b^{2}-3ac)^{3}}}}}=\cos {3\theta }={\frac {e^{3i\theta }+e^{-3i\theta }}{2}}{\text{ yielding 3 real solutions:}}$ $x_{k}={\frac {\left(e^{i\left(\theta +{\frac {2k\pi }{3}}\right)}+e^{-i\left(\theta +{\frac {2k\pi }{3}}\right)}\right){\sqrt {b^{2}-3ac}}-b}{3a}}=2\cos \left({\frac {1}{3}}\arccos {\frac {9abc-2b^{3}-27a^{2}d}{2{\sqrt {(b^{2}-3ac)^{3}}}}}+{\frac {2k\pi }{3}}\right){\frac {\sqrt {b^{2}-3ac}}{3a}}-{\frac {b}{3a}}.$

Last five lines are presenting all that the innocent little children should know about the resolving of cubic equation – the memorizing of the items involved is facilitated hence all of three are either equal or proportional to the inflection point properties. See the examples: $x_{0}={\frac {1}{2}}\left({\sqrt[{3}]{7+{\sqrt {7^{2}+1^{3}}}}}+{\sqrt[{3}]{7-{\sqrt {7^{2}+1^{3}}}}}\right)=\sinh {\frac {\operatorname {arsinh} {7}}{3}}=1{\text{ is real solution of }}4x^{3}+3x=7{\text{ and}}$ $x_{0}={\frac {1}{2}}\left({\sqrt[{3}]{26+{\sqrt {26^{2}-1^{3}}}}}+{\sqrt[{3}]{26-{\sqrt {26^{2}-1^{3}}}}}\right)=\cosh {\frac {\operatorname {arcosh} {26}}{3}}=2{\text{ is real solution of }}4x^{3}-3x=26{\text{ but}}$$x_{k}=2*7\cos \left({\frac {1}{3}}\arccos {{\frac {-286}{2*7^{3}}}+{\frac {2k\pi }{3}}}\right)+1=12;-12;3{\text{ are real solutions of }}(x-1)^{3}-3*7^{2}(x-1)=-286.$ Note: Obviously the evaluation of first two examples will be much easier if hyperbolical formulae would be applied.

Comment: It’s incomprehensible that a point of such importance for THE FLOW OF THE FUNCTION isn’t mentioned within entire article although the coordinates and slope of S are determining all remaining characteristic points: the zero(s), critical points, if any, as well. Moreover, the circle at Figure 4 is unnecessary dislocated above S(x_S,y_S) omitting even a vertical line up to the center of the circle as done at Figure 2 of “New approach …” by Nickalls R. W. D. (see note 29 again). Tschirnhaus transformation isn’t only unneeded but also responsible for such a failure hence t_S = 0 doesn’t mean that an inflection point S(0,q) doesn’t exist. It seems to me reasonable the unknown t along with p and q to be abandoned in favour of z, θ and geometrically grounded variables which are also easier for memorizing.

Note: Letter S (instead N at 29) is chosen not only for Symmetry but also due to its shape reminding to cubic function.

217.197.142.0 (talk) 09:02, 19 July 2013 (UTC)Stap Modified many times by 217.197.142.0 (talk) 23:13, 2 May 2014 (UTC)