Editing Cubic equation (section)

==Derivation of the roots==
This section regroups several methods for deriving [[#Cardano's formula|Cardano's formula]].

===Cardano's method===
This method is due to [[Scipione del Ferro]] and [[Niccolò Fontana Tartaglia|Tartaglia]], but is named after [[Gerolamo Cardano]] who first published it in his book [[Ars Magna (Gerolamo Cardano)|''Ars Magna'']] (1545).

This method applies to a depressed cubic {{math|''t''<sup>3</sup> + ''pt'' + ''q'' {{=}} 0}}. The idea is to introduce two variables {{mvar|u}} and <math>v</math> such that <math>u+v=t</math> and to substitute this in the depressed cubic, giving
<math display="block">u^3 + v^3 + (3uv + p)(u+v)+ q= 0.</math>

At this point Cardano imposed the condition <math>3uv+p=0.</math> This removes the third term in the previous equality, leading to the system of equations
<math display="block">\begin{align}u^3 + v^3&=-q \\ uv&=-\frac p3.\end{align}</math>

Knowing the sum and the product of {{math|''u''<sup>3</sup>}} and <math>v^3,</math> one deduces that they are the two solutions of the [[quadratic equation]] 
<math display="block">\begin{align}
  0 &= (x - u^3)(x - v^3) \\ 
  &= x^2 - (u^3 + v^3)x + u^3v^3 \\
  &= x^2 - (u^3 + v^3)x + (uv)^3
 \end{align}</math> 
so
<math display="block">x^2 + qx -\frac {p^3}{27}=0.</math>
The discriminant of this equation is <math>\Delta = q^2 + \frac{4p^3}{27}</math>, and assuming it is positive, real solutions to this equation are (after folding division by 4 under the square root):
<math display="block">-\frac q2 \pm \sqrt {\frac {q^2}{4} +\frac {p^3}{27}}.</math>
So (without loss of generality in choosing {{mvar|u}} or <math>v</math>):
<math display="block">u = \sqrt[3]{-\frac q2 + \sqrt {\frac {q^2}{4} +\frac {p^3}{27}}}.</math>
<math display="block">v = \sqrt[3]{-\frac q2 - \sqrt {\frac {q^2}{4} +\frac {p^3}{27}}}.</math>
As <math>u+v=t,</math> the sum of the cube roots of these solutions is a root of the equation. That is 
<math display="block">t=\sqrt[3]{-{q\over 2}+\sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2}-\sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}</math>
is a root of the equation; this is Cardano's formula.

This works well when <math>4p^3+27q^2 > 0,</math> but, if <math>4p^3+27q^2 < 0,</math> the square root appearing in the formula is not real. As a [[complex number]] has three cube roots, using Cardano's formula without care would provide nine roots, while a cubic equation cannot have more than three roots. This was clarified first by [[Rafael Bombelli]] in his book ''L'Algebra'' (1572). The solution is to use the fact that <math>uv=-\frac p3,</math> that is, <math>v=\frac{-p}{3u}.</math> This means that only one cube root needs to be computed, and leads to the second formula given in {{slink||Cardano's formula}}.

The other roots of the equation can be obtained by changing of cube root, or, equivalently, by multiplying the cube root by each of the two [[primitive root of unity|primitive cube roots of unity]], which are <math>\frac {-1\pm \sqrt{-3}}2.</math>

===Vieta's substitution===
Vieta's substitution is a method introduced by [[François Viète]] (Vieta is his Latin name) in a text published posthumously in 1615, which provides directly the second formula of {{slink||Cardano's method}}, and avoids the problem of computing two different cube roots.<ref>{{Citation|last = van der Waerden|first = Bartel Leenert|author-link = Bartel Leendert van der Waerden|title = A History of Algebra: From al-Khwārizmī to Emmy Noether|publisher = [[Springer Science+Business Media|Springer-Verlag]]|year = 1985|chapter = From Viète to Descartes|isbn = 3-540-13610-X}}</ref>

Starting from the depressed cubic {{math|''t''<sup>3</sup> + ''pt'' + ''q'' {{=}} 0}}, Vieta's substitution is {{math|''t'' {{=}} ''w'' − {{sfrac|''p''|3''w''}}}}.{{efn|More precisely, Vieta introduced a new variable {{mvar|w}} and imposed the condition {{math|''w''(''t'' + ''w'') {{=}} {{sfrac|''p''|3}}}}. This is equivalent with the substitution {{math|''t'' {{=}} {{sfrac|''p''|3''w''}} − ''w''}}, and differs from the substitution that is used here only by a change of sign of {{mvar|w}}. This change of sign allows getting directly the formulas of {{slink||Cardano's formula}}.}}

The substitution {{math|''t'' {{=}} ''w'' – {{sfrac|''p''|3''w''}}}} transforms the depressed cubic into
<math display="block">w^3+q-\frac{p^3}{27w^3}=0.</math>

Multiplying by {{math|''w''<sup>3</sup>}}, one gets a quadratic equation in {{mvar|w{{sup|3}}}}:
<math display="block">(w^3)^2+q(w^3)-\frac{p^3}{27}=0.</math>

Let 
<math display="block">W=-\frac q 2\pm\sqrt{\frac{p^3}{27} + \frac {q^2} 4}</math> 
be any nonzero root of this quadratic equation. If {{math|''w''<sub>1</sub>}}, {{math|''w''<sub>2</sub>}} and {{math|''w''<sub>3</sub>}} are the three [[cube root]]s of {{mvar|W}}, then the roots of the original depressed cubic are {{math|''w''<sub>1</sub> − {{sfrac|''p''|3''w''<sub>1</sub>}}}}, {{math|''w''<sub>2</sub> − {{sfrac|''p''|3''w''<sub>2</sub>}}}}, and {{math|''w''<sub>3</sub> − {{sfrac|''p''|3''w''<sub>3</sub>}}}}. The other root of the quadratic equation is <math>\textstyle -\frac {p^3}{27W}.</math> This implies that changing the sign of the square root exchanges {{math|''w''<sub>''i''</sub>}} and {{math|− {{sfrac|''p''|3''w''<sub>''i''</sub>}}}} for {{math|1=''i'' = 1, 2, 3}}, and therefore does not change the roots. This method only fails when both roots of the quadratic equation are zero, that is when {{math|''p'' {{=}} ''q'' {{=}} 0}}, in which case the only root of the depressed cubic is {{math|0}}.

===Lagrange's method===
In his paper ''Réflexions sur la résolution algébrique des équations'' ("Thoughts on the algebraic solving of equations"),<ref>{{Citation|last = Lagrange|first = Joseph-Louis|author-link = Joseph Louis Lagrange|chapter = Réflexions sur la résolution algébrique des équations|title = Œuvres de Lagrange|editor-last = Serret|editor-first = Joseph-Alfred|editor-link = Joseph Alfred Serret|year = 1869|orig-year = 1771|publisher = Gauthier-Villars|pages = 205–421|volume = III}}</ref> [[Joseph Louis Lagrange]] introduced a new method to solve equations of low degree in a uniform way, with the hope that he could generalize it for higher degrees. This method works well for cubic and [[quartic equation]]s, but Lagrange did not succeed in applying it to a [[quintic equation]], because it requires solving a resolvent polynomial of degree at least six.<ref name="efei">{{citation
|title=Elliptic functions and elliptic integrals
|first1=Viktor
|last1=Prasolov
|first2=Yuri
|last2=Solovyev
|publisher=AMS Bookstore
|year=1997
|isbn=978-0-8218-0587-9
|url=https://books.google.com/books?id=fcp9IiZd3tQC
}}, [https://books.google.com/books?id=fcp9IiZd3tQC&pg=PA134 §6.2, p. 134]</ref><ref>{{citation
|first=Morris
|last=Kline
|title=Mathematical Thought from Ancient to Modern Times
|publisher=Oxford University Press US
|year=1990
|isbn=978-0-19-506136-9
|url=https://books.google.com/books?id=aO-v3gvY-I8C
}}, [https://books.google.com/books?id=aO-v3gvY-I8C Algebra in the Eighteenth Century: The Theory of Equations]</ref><ref name="laz">Daniel Lazard, "Solving quintics in radicals", in [[Olav Arnfinn Laudal]], [[Ragni Piene]], ''The Legacy of Niels Henrik Abel'', pp.&nbsp;207–225, Berlin, 2004. {{isbn|3-540-43826-2}}</ref> 
Apart from the fact that nobody had previously succeeded, this was the first indication of the non-existence of an algebraic formula for degrees 5 and higher; as was later proved by the [[Abel–Ruffini theorem]]. Nevertheless, modern methods for solving solvable quintic equations are mainly based on Lagrange's method.<ref name="laz" />

In the case of cubic equations, Lagrange's method gives the same solution as Cardano's. Lagrange's method can be applied directly to the general cubic equation  {{math|''ax''<sup>3</sup> + ''bx''<sup>2</sup> + ''cx'' + ''d'' {{=}} 0}}, but the computation is simpler with the depressed cubic equation, {{math|''t''<sup>3</sup> + ''pt'' + ''q'' {{=}} 0}}.

Lagrange's main idea was to work with the [[discrete Fourier transform]] of the roots instead of with the roots themselves. More precisely, let {{mvar|ξ}} be a [[primitive root of unity|primitive third root of unity]], that is a number such that {{math|''ξ''<sup>3</sup> {{=}} 1}} and {{math|''ξ''<sup>2</sup> + ''ξ'' + 1 {{=}} 0}} (when working in the space of [[complex number]]s, one has <math>\textstyle \xi=\frac{-1\pm i\sqrt 3}2=e^{2i\pi/3},</math> but this complex interpretation is not used here). Denoting {{math|''x''<sub>0</sub>}}, {{math|''x''<sub>1</sub>}} and {{math|''x''<sub>2</sub>}} the three roots of the cubic equation to be solved, let
<math display="block">\begin{align}
s_0 &= x_0 + x_1 + x_2,\\
s_1 &= x_0 + \xi x_1 + \xi^2 x_2,\\
s_2 &= x_0 + \xi^2 x_1 + \xi x_2,
\end{align}</math>
be the discrete Fourier transform of the roots. If {{math|''s''<sub>0</sub>}}, {{math|''s''<sub>1</sub>}} and {{math|''s''<sub>2</sub>}} are known, the roots may be recovered from them with the inverse Fourier transform consisting of inverting this linear transformation; that is,
<math display="block">\begin{align}
x_0 &= \tfrac13(s_0 + s_1 + s_2),\\
x_1 &= \tfrac13(s_0 + \xi^2 s_1 + \xi s_2),\\
x_2 &= \tfrac13(s_0 + \xi s_1 + \xi ^2 s_2).
\end{align}</math>

By [[Vieta's formulas]], {{math|''s''<sub>0</sub>}} is known to be zero in the case of a depressed cubic, and {{math|−{{sfrac|''b''|''a''}}}} for the general cubic. So, only {{math|''s''<sub>1</sub>}} and {{math|''s''<sub>2</sub>}} need to be computed. They are not [[symmetric polynomial|symmetric functions]] of the roots (exchanging {{math|''x''<sub>1</sub>}} and {{math|''x''<sub>2</sub>}} exchanges also {{math|''s''<sub>1</sub>}} and {{math|''s''<sub>2</sub>}}), but some simple symmetric functions of {{math|''s''<sub>1</sub>}} and {{math|''s''<sub>2</sub>}} are also symmetric in the roots of the cubic equation to be solved. Thus these symmetric functions can be expressed in terms of the (known) coefficients of the original cubic, and this allows eventually expressing the {{mvar|s{{sub|i}}}} as roots of a polynomial with known coefficients. This works well for every degree, but, in degrees higher than four, the resulting polynomial that has the {{mvar|s{{sub|i}}}} as roots has a degree higher than that of the initial polynomial, and is therefore unhelpful for solving. This is the reason for which Lagrange's method fails in degrees five and higher.

In the case of a cubic equation, <math>P=s_1s_2,</math> and <math>S=s_1^3+s_2^3</math> are such symmetric polynomials (see below). It follows that <math>s_1^3</math> and <math>s_2^3</math> are the two roots of the quadratic equation <math>z^2-Sz+P^3=0.</math> Thus the resolution of the equation may be finished exactly as with Cardano's method, with <math>s_1</math> and <math>s_2</math> in place of {{mvar|u}} and <math>v.</math>

In the case of the depressed cubic, one has <math>x_0=\tfrac 13 (s_1+s_2)</math> and <math>s_1s_2=-3p,</math> while in Cardano's method we have set <math>x_0=u+v</math> and <math>uv=-\tfrac 13 p.</math> Thus, up to the exchange of {{mvar|u}} and <math>v,</math> we have <math>s_1=3u</math> and <math>s_2=3v.</math> In other words, in this case, Cardano's method and Lagrange's method compute exactly the same things, up to a factor of three in the auxiliary variables, the main difference being that Lagrange's method explains why these auxiliary variables appear in the problem.

====Computation of {{mvar|S}} and {{mvar|P}}====
A straightforward computation using the relations {{math|''ξ''<sup>3</sup> {{=}} 1}} and {{math|''ξ''<sup>2</sup> + ''ξ'' + 1 {{=}} 0}} gives 
<math display="block">\begin{align}
P&=s_1s_2=x_0^2+x_1^2+x_2^2-(x_0x_1+x_1x_2+x_2x_0),\\
S&=s_1^3+s_2^3=2(x_0^3+x_1^3+x_2^3)-3(x_0^2x_1+x_1^2x_2+x_2^2x_0+x_0x_1^2+x_1x_2^2+x_2x_0^2)+12x_0x_1x_2.
\end{align}</math>
This shows that {{mvar|P}} and {{mvar|S}} are symmetric functions of the roots. Using [[Newton's identities]], it is straightforward to express them in terms of the [[elementary symmetric polynomial|elementary symmetric functions]] of the roots, giving
<math display="block">\begin{align}
P&=e_1^2-3e_2,\\
S&=2e_1^3-9e_1e_2+27e_3,
\end{align}</math>
with {{math|''e''<sub>1</sub> {{=}} 0}}, {{math|''e''<sub>2</sub> {{=}} ''p''}} and {{math|''e''<sub>3</sub> {{=}} −''q''}} in the case of a depressed cubic, and {{math|''e''<sub>1</sub> {{=}} −{{sfrac|''b''|''a''}}}}, {{math|''e''<sub>2</sub> {{=}} {{sfrac|''c''|''a''}}}} and {{math|''e''<sub>3</sub> {{=}} −{{sfrac|''d''|''a''}}}}, in the general case.