Graph of a cubic function with 3realroots (where the curve crosses the horizontal axis aty = 0). The case shown has twocritical points. Here the function is and therefore the three real roots are 2, −1 and −4.
Inalgebra, acubic equation in one variable is anequation of the formin whicha is not zero.
The solutions of this equation are calledroots of thecubic function defined by the left-hand side of the equation. If all of thecoefficientsa,b,c, andd of the cubic equation arereal numbers, then it has at least one real root (this is true for all odd-degreepolynomial functions). All of the roots of the cubic equation can be found by the following means:
The coefficients do not need to be real numbers. Much of what is covered below is valid for coefficients in anyfield withcharacteristic other than 2 and 3. The solutions of the cubic equation do not necessarily belong to the same field as the coefficients. For example, some cubic equations with rational coefficients have roots that are irrational (and even non-real)complex numbers.
Cubic equations were known to the ancient Babylonians, Greeks, Chinese, Indians, and Egyptians.[1][2][3]Babylonian (20th to 16th centuries BC) cuneiform tablets have been found with tables for calculating cubes and cube roots.[4][5] The Babylonians could have used the tables to solve cubic equations, but no evidence exists to confirm that they did.[6] The problem ofdoubling the cube involves the simplest and oldest studied cubic equation, and one for which the ancient Egyptians did not believe a solution existed.[7] In the 5th century BC,Hippocrates reduced this problem to that of finding two mean proportionals between one line and another of twice its length, but could not solve this with acompass and straightedge construction,[8] a task which is now known to be impossible. Methods for solving cubic equations appear inThe Nine Chapters on the Mathematical Art, aChinese mathematical text compiled around the 2nd century BC and commented on byLiu Hui in the 3rd century.[2]
In the 3rd century AD, theGreek mathematicianDiophantus found integer or rational solutions for some bivariate cubic equations (Diophantine equations).[3][9] Hippocrates,Menaechmus andArchimedes are believed to have come close to solving the problem of doubling the cube using intersectingconic sections,[8] though historians such as Reviel Netz dispute whether the Greeks were thinking about cubic equations or just problems that can lead to cubic equations. Some others likeT. L. Heath, who translated all of Archimedes's works, disagree, putting forward evidence that Archimedes really solved cubic equations using intersections of twoconics, but also discussed the conditions where the roots are 0, 1 or 2.[10]
In the 7th century, theTang dynasty astronomer mathematicianWang Xiaotong in his mathematical treatise titledJigu Suanjing systematically established and solvednumerically 25 cubic equations of the formx3 +px2 +qx =N, 23 of them withp,q ≠ 0, and two of them withq = 0.[11]
In the 11th century, the Persian poet-mathematician,Omar Khayyam (1048–1131), made significant progress in the theory of cubic equations. In an early paper, he discovered that a cubic equation can have more than one solution and stated that it cannot be solved using compass and straightedge constructions. He also found a geometric solution.[12][a] In his later work, theTreatise on Demonstration of Problems of Algebra, he wrote a complete classification of cubic equations with general geometric solutions found by means of intersectingconic sections.[13][14] Khayyam made an attempt to come up with an algebraic formula for extracting cubic roots. He wrote:
“We have tried to express these roots by algebra but have failed. It may be, however, that men who come after us will succeed.”[15]
In the 12th century, the Indian mathematician Bhaskara II attempted the solution of cubic equations without general success. However, he gave one example of a cubic equation:x3 + 12x = 6x2 + 35.[16] In the 12th century, anotherPersian mathematician,Sharaf al-Dīn al-Tūsī (1135–1213), wrote theAl-Muʿādalāt (Treatise on Equations), which dealt with eight types of cubic equations with positive solutions and five types of cubic equations which may not have positive solutions. He used what would later be known as theHorner–Ruffini method tonumerically approximate theroot of a cubic equation. He also used the concepts ofmaxima and minima of curves in order to solve cubic equations which may not have positive solutions.[17] He understood the importance of thediscriminant of the cubic equation to find algebraic solutions to certain types of cubic equations.[18]
In his bookFlos, Leonardo de Pisa, also known asFibonacci (1170–1250), was able to closely approximate the positive solution to the cubic equationx3 + 2x2 + 10x = 20. Writing inBabylonian numerals he gave the result as 1,22,7,42,33,4,40 (equivalent to 1 + 22/60 + 7/602 + 42/603 + 33/604 + 4/605 + 40/606), which has arelative error of about 10−9.[19]
In the early 16th century, the Italian mathematicianScipione del Ferro (1465–1526) found a method for solving a class of cubic equations, namely those of the formx3 +mx =n. In fact, all cubic equations can be reduced to this form if one allowsm andn to be negative, butnegative numbers were not known to him at that time. Del Ferro kept his achievement secret until just before his death, when he told his student Antonio Fior about it.
Niccolò Fontana Tartaglia
In 1535,Niccolò Tartaglia (1500–1557) received two problems in cubic equations fromZuanne da Coi and announced that he could solve them. He was soon challenged by Fior, which led to a famous contest between the two. Each contestant had to put up a certain amount of money and to propose a number of problems for his rival to solve. Whoever solved more problems within 30 days would get all the money. Tartaglia received questions in the formx3 +mx =n, for which he had worked out a general method. Fior received questions in the formx3 +mx2 =n, which proved to be too difficult for him to solve, and Tartaglia won the contest.
Later, Tartaglia was persuaded byGerolamo Cardano (1501–1576) to reveal his secret for solving cubic equations. In 1539, Tartaglia did so only on the condition that Cardano would never reveal it and that if he did write a book about cubics, he would give Tartaglia time to publish. Some years later, Cardano learned about del Ferro's prior work and published del Ferro's method in his bookArs Magna in 1545, meaning Cardano gave Tartaglia six years to publish his results (with credit given to Tartaglia for an independent solution).
Cardano's promise to Tartaglia said that he would not publish Tartaglia's work, and Cardano felt he was publishing del Ferro's, so as to get around the promise. Nevertheless, this led to a challenge to Cardano from Tartaglia, which Cardano denied. The challenge was eventually accepted by Cardano's studentLodovico Ferrari (1522–1565). Ferrari did better than Tartaglia in the competition, and Tartaglia lost both his prestige and his income.[20]
Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with thesecomplex numbers inArs Magna, but he did not really understand it.Rafael Bombelli studied this issue in detail[21] and is therefore often considered as the discoverer of complex numbers.
François Viète (1540–1603) independently derived the trigonometric solution for the cubic with three real roots, andRené Descartes (1596–1650) extended the work of Viète.[22]
If the coefficients of a cubic equation arerational numbers, one can obtain an equivalent equation with integer coefficients, by multiplying all coefficients by acommon multiple of their denominators. Such an equationwith integer coefficients, is said to bereducible if the polynomial on the left-hand side is the product of polynomials of lower degrees. ByGauss's lemma, if the equation is reducible, one can suppose that thefactors have integer coefficients.
Finding the roots of a reducible cubic equation is easier than solving the general case. In fact, if the equation is reducible, one of the factors must have degree one, and thus have the formwithq andp beingcoprime integers. Therational root test allows findingq andp by examining a finite number of cases (becauseq must be a divisor ofa, andp must be a divisor ofd).
Thus, one root is and the other roots are the roots of the other factor, which can be found bypolynomial long division. This other factor is(The coefficients seem not to be integers, but must be integers if is a root.)
Cubics of the formare said to be depressed. They are much simpler than general cubics, but are fundamental, because the study of any cubic may be reduced by a simplechange of variable to that of a depressed cubic.
Letbe a cubic equation. The change of variablegives a cubic (int) that has no term int2.
After dividing bya one gets thedepressed cubic equationwith
Theroots of the original equation are related to the roots of the depressed equation by the relationsfor.
Thediscriminant of apolynomial is a function of its coefficients that is zero if and only if the polynomial has amultiple root, or, if it is divisible by the square of a non-constant polynomial. In other words, the discriminant is nonzero if and only if the polynomial issquare-free.
Ifr1,r2,r3 are the threeroots (not necessarily distinct norreal) of the cubic then the discriminant is
The discriminant of the depressed cubic is
The discriminant of the general cubic isIt is the product of and the discriminant of the corresponding depressed cubic. Using the formula relating the general cubic and the associated depressed cubic, this implies that the discriminant of the general cubic can be written as
It follows that one of these two discriminants is zero if and only if the other is also zero, and, if the coefficients arereal, the two discriminants have the same sign. In summary, the same information can be deduced from either one of these two discriminants.
To prove the preceding formulas, one can useVieta's formulas to express everything as polynomials inr1,r2,r3, anda. The proof then results in the verification of the equality of two polynomials.
If the cubic has one real root and two non-realcomplex conjugate roots.
This can be proved as follows. First, ifr is a root of a polynomial with real coefficients, then itscomplex conjugate is also a root. So the non-real roots, if any, occur as pairs of complex conjugate roots. As a cubic polynomial has three roots (not necessarily distinct) by thefundamental theorem of algebra, at least one root must be real.
As stated above, ifr1,r2,r3 are the three roots of the cubic, then the discriminant is
If the three roots are real and distinct, the discriminant is a product of positive reals, that is
If only one root, sayr1, is real, thenr2 andr3 are complex conjugates, which implies thatr2 −r3 is apurely imaginary number, and thus that(r2 −r3)2 is real and negative. On the other hand,r1 −r2 andr1 −r3 are complex conjugates, and their product is real and positive.[23] Thus the discriminant is the product of a single negative number and several positive ones. That is
If the discriminant of a cubic is zero, the cubic has amultiple root. If furthermore its coefficients are real, then all of its roots are real.
The discriminant of the depressed cubic is zero if Ifp is also zero, thenp =q = 0, and 0 is a triple root of the cubic. If andp ≠ 0, then the cubic has a simple root
and a double root
In other words,
This result can be proved by expanding the latter product or retrieved by solving the rather simplesystem of equations resulting fromVieta's formulas.
By using thereduction of a depressed cubic, these results can be extended to the general cubic. This gives: If the discriminant of the cubic is zero, then
either, if the cubic has a triple root and
or, if the cubic has a double root and a simple root, and thus
The above results are valid when the coefficients belong to afield ofcharacteristic other than 2 or 3, but must be modified for characteristic 2 or 3, because of the involved divisions by 2 and 3.
The reduction to a depressed cubic works for characteristic 2, but not for characteristic 3. However, in both cases, it is simpler to establish and state the results for the general cubic. The main tool for that is the fact that a multiple root is a common root of the polynomial and itsformal derivative. In these characteristics, if the derivative is not a constant, it is a linear polynomial in characteristic 3, and is the square of a linear polynomial in characteristic 2. Therefore, for either characteristic 2 or 3, the derivative has only one root. This allows computing the multiple root, and the third root can be deduced from the sum of the roots, which is provided byVieta's formulas.
A difference with other characteristics is that, in characteristic 2, the formula for a double root involves a square root, and, in characteristic 3, the formula for a triple root involves a cube root.
Cardano's result is that ifis a cubic equation such thatp andq arereal numbers such that is positive (this implies that thediscriminant of the equation is negative) then the equation has the real rootwhere and are the two numbers and
As shown in§ Nature of the roots, the two other roots are non-realcomplex conjugate numbers, in this case. It was later shown (Cardano did not knowcomplex numbers) that the two other roots are obtained by multiplying one of the cube roots by theprimitive cube root of unity and the other cube root by the other primitive cube root of the unity That is, the other roots of the equation are and[24]
If there are three real roots, butGalois theory allows proving that, if there is no rational root, the roots cannot be expressed by analgebraic expression involving only real numbers. Therefore, the equation cannot be solved in this case with the knowledge of Cardano's time. This case has thus been calledcasus irreducibilis, meaningirreducible case in Latin.
Incasus irreducibilis, Cardano's formula can still be used, but some care is needed in the use of cube roots. A first method is to define the symbols and as representing theprincipal values of the root function (that is the root that has the largest real part). With this convention Cardano's formula for the three roots remains valid, but is not purely algebraic, as the definition of a principal part is not purely algebraic, since it involves inequalities for comparing real parts. Also, the use of principal cube root may give a wrong result if the coefficients are non-real complex numbers. Moreover, if the coefficients belong to anotherfield, the principal cube root is not defined in general.
The second way for making Cardano's formula always correct, is to remark that the product of the two cube roots must be−p / 3. It results that a root of the equation isIn this formula, the symbols and denote any square root and any cube root. The other roots of the equation are obtained either by changing of cube root or, equivalently, by multiplying the cube root by a primitive cube root of unity, that is
This formula for the roots is always correct except whenp =q = 0, with the proviso that ifp = 0, the square root is chosen so thatC ≠ 0. However, Cardano's formula is useless if as the roots are the cube roots of Similarly, the formula is also useless in the cases where no cube root is needed, that is when the cubic polynomial is notirreducible; this includes the case
This formula is also correct whenp andq belong to anyfield ofcharacteristic other than 2 or 3.
Acubic formula for the roots of the general cubic equation (witha ≠ 0)can be deduced from every variant of Cardano's formula by reduction to adepressed cubic. The variant that is presented here is valid not only for complex coefficients, but also for coefficientsa,b,c,d belonging to anyalgebraically closed field ofcharacteristic other than 2 or 3. If the coefficients are real numbers, the formula covers all complex solutions, not just real ones.
The formula being rather complicated, it is worth splitting it in smaller formulas.
Let
(Both and can be expressed asresultants of the cubic and its derivatives: is−1/8a times the resultant of the cubic and its second derivative, and is−1/12a times the resultant of the first and second derivatives of the cubic polynomial.)
Then letwhere the symbols and are interpreted asany square root andany cube root, respectively (every nonzero complex number has two square roots and three cubic roots). The sign "±" before the square root is either "+" or "–"; the choice is almost arbitrary, and changing it amounts to choosing a different square root. However, if a choice yieldsC = 0 (this occurs if), then the other sign must be selected instead. If both choices yieldC = 0, that is, if a fraction0/0 occurs in following formulas; this fraction must be interpreted as equal to zero (see the end of this section).With these conventions, one of the roots is
The other two roots can be obtained by changing the choice of the cube root in the definition ofC, or, equivalently by multiplyingC by aprimitive cube root of unity, that is–1 ±√–3/2. In other words, the three roots arewhereξ =–1 +√–3/2.
As for the special case of a depressed cubic, this formula applies but is useless when the roots can be expressed without cube roots. In particular, if the formula gives that the three roots equal which means that the cubic polynomial can be factored as A straightforward computation allows verifying that the existence of this factorization is equivalent with
When a cubic equation with real coefficients has three real roots, the formulas expressing these roots in terms of radicals involve complex numbers.Galois theory allows proving that when the three roots are real, and none is rational (casus irreducibilis), one cannot express the roots in terms of real radicals. Nevertheless, purely real expressions of the solutions may be obtained usingtrigonometric functions, specifically in terms ofcosines andarccosines.[25] More precisely, the roots of thedepressed cubicare[26]
This formula is due toFrançois Viète.[22] It is purely real when the equation has three real roots (that is). Otherwise, it is still correct but involves complex cosines and arccosines when there is only one real root, and it is nonsensical (division by zero) whenp = 0.
This formula can be straightforwardly transformed into a formula for the roots of a general cubic equation, using the back-substitution described in§ Depressed cubic.
The formula can be proved as follows: Starting from the equationt3 +pt +q = 0, let us sett =u cos θ. The idea is to chooseu to make the equation coincide with the identityFor this, choose and divide the equation by This givesCombining with the above identity, one getsand the roots are thus
When there is only one real root (andp ≠ 0), this root can be similarly represented usinghyperbolic functions, as[27][28]Ifp ≠ 0 and the inequalities on the right are not satisfied (the case of three real roots), the formulas remain valid but involve complex quantities.
Whenp = ±3, the above values oft0 are sometimes called theChebyshev cube root.[29] More precisely, the values involving cosines and hyperbolic cosines define, whenp = −3, the sameanalytic function denotedC1/3(q), which is the proper Chebyshev cube root. The value involving hyperbolic sines is similarly denotedS1/3(q), whenp = 3.
Omar Khayyám's geometric solution of a cubic equation, for the casem = 2,n = 16, giving the root2. The intersection of the vertical line on thex-axis at the center of the circle is happenstance of the example illustrated.
For solving the cubic equationx3 +m2x =n wheren > 0,Omar Khayyám constructed the parabolay =x2/m, the circle that has as a diameter theline segment[0,n/m2] on the positivex-axis, and a vertical line through the point where the circle and the parabola intersect above thex-axis. The solution is given by the length of the horizontal line segment from the origin to the intersection of the vertical line and thex-axis (see the figure).
A simple modern proof is as follows. Multiplying the equation byx/m2 and regrouping the terms givesThe left-hand side is the value ofy2 on the parabola. The equation of the circle beingy2 +x(x −n/m2) = 0, the right hand side is the value ofy2 on the circle.
A cubic equation with real coefficients can be solved geometrically usingcompass, straightedge, and anangle trisector if and only if it has three real roots.[30]: Thm. 1
A cubic equation can be solved by compass-and-straightedge construction (without trisector) if and only if it has arational root. This implies that the old problems ofangle trisection anddoubling the cube, set byancient Greek mathematicians, cannot be solved by compass-and-straightedge construction.
For the cubic(1) with three real roots, the roots are the projection on thex-axis of the verticesA,B, andC of anequilateral triangle. The center of the triangle has the samex-coordinate as theinflection point.
Viète's trigonometric expression of the roots in the three-real-roots case lends itself to a geometric interpretation in terms of a circle.[22][31] When the cubic is written in depressed form(2),t3 +pt +q = 0, as shown above, the solution can be expressed as
Here is an angle in the unit circle; taking1/3 of that angle corresponds to taking a cube root of a complex number; adding−k2π/3 fork = 1, 2 finds the other cube roots; and multiplying the cosines of these resulting angles by corrects for scale.
For the non-depressed case(1) (shown in the accompanying graph), the depressed case as indicated previously is obtained by definingt such thatx =t −b/3a sot =x +b/3a. Graphically this corresponds to simply shifting the graph horizontally when changing between the variablest andx, without changing the angle relationships. This shift moves the point of inflection and the centre of the circle onto they-axis. Consequently, the roots of the equation int sum to zero.
The slope of line RA is twice that of RH. Denoting the complex roots of the cubic asg ±hi,g =OM (negative here) andh =√tanORH =√slope of lineRH =BE =DA.
When the graph of acubic function is plotted in theCartesian plane, if there is only one real root, it is theabscissa (x-coordinate) of the horizontal intercept of the curve (point R on the figure). Further,[32][33][34] if the complex conjugate roots are written asg ±hi, then thereal partg is the abscissa of the tangency point H of thetangent line to cubic that passes throughx-intercept R of the cubic (that is the signed length OM, negative on the figure). Theimaginary parts±h are the square roots of the tangent of the angle between this tangent line and the horizontal axis.[clarification needed]
With one real and two complex roots, the three roots can be represented as points in the complex plane, as can the two roots of the cubic's derivative. There is an interesting geometrical relationship among all these roots.
The points in the complex plane representing the three roots serve as the vertices of an isosceles triangle. (The triangle is isosceles because one root is on the horizontal (real) axis and the other two roots, being complex conjugates, appear symmetrically above and below the real axis.)Marden's theorem says that the points representing the roots of the derivative of the cubic are thefoci of theSteiner inellipse of the triangle—the unique ellipse that is tangent to the triangle at the midpoints of its sides. If the angle at the vertex on the real axis is less thanπ/3 then the major axis of the ellipse lies on the real axis, as do its foci and hence the roots of the derivative. If that angle is greater thanπ/3, the major axis is vertical and its foci, the roots of the derivative, are complex conjugates. And if that angle isπ/3, the triangle is equilateral, the Steiner inellipse is simply the triangle's incircle, its foci coincide with each other at the incenter, which lies on the real axis, and hence the derivative has duplicate real roots.
Given a cubicirreducible polynomial over a fieldK ofcharacteristic different from 2 and 3, theGalois group overK is the group of thefield automorphisms that fixK of the smallest extension ofK (splitting field). As these automorphisms must permute the roots of the polynomials, this group is either the groupS3 of all six permutations of the three roots, or the groupA3 of the three circular permutations.
The discriminantΔ of the cubic is the square ofwherea is the leading coefficient of the cubic, andr1,r2 andr3 are the three roots of the cubic. As changes of sign if two roots are exchanged, is fixed by the Galois group only if the Galois group isA3. In other words, the Galois group isA3 if and only if the discriminant is the square of an element ofK.
As most integers are not squares, when working over the fieldQ of therational numbers, the Galois group of most irreducible cubic polynomials is the groupS3 with six elements. An example of a Galois groupA3 with three elements is given byp(x) =x3 − 3x − 1, whose discriminant is81 = 92.
This method applies to a depressed cubict3 +pt +q = 0. The idea is to introduce two variablesu and such that and to substitute this in the depressed cubic, giving
At this point Cardano imposed the condition This removes the third term in the previous equality, leading to the system of equations
Knowing the sum and the product ofu3 and one deduces that they are the two solutions of thequadratic equation soThe discriminant of this equation is, and assuming it is positive, real solutions to this equation are (after folding division by 4 under the square root):So (without loss of generality in choosingu or):As the sum of the cube roots of these solutions is a root of the equation. That isis a root of the equation; this is Cardano's formula.
This works well when but, if the square root appearing in the formula is not real. As acomplex 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 byRafael Bombelli in his bookL'Algebra (1572). The solution is to use the fact that that is, This means that only one cube root needs to be computed, and leads to the second formula given in§ 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 twoprimitive cube roots of unity, which are
Vieta's substitution is a method introduced byFrançois Viète (Vieta is his Latin name) in a text published posthumously in 1615, which provides directly the second formula of§ Cardano's method, and avoids the problem of computing two different cube roots.[35]
Starting from the depressed cubict3 +pt +q = 0, Vieta's substitution ist =w −p/3w.[b]
The substitutiont =w –p/3w transforms the depressed cubic into
Multiplying byw3, one gets a quadratic equation inw3:
Let be any nonzero root of this quadratic equation. Ifw1,w2 andw3 are the threecube roots ofW, then the roots of the original depressed cubic arew1 −p/3w1,w2 −p/3w2, andw3 −p/3w3. The other root of the quadratic equation is This implies that changing the sign of the square root exchangeswi and−p/3wi fori = 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 whenp =q = 0, in which case the only root of the depressed cubic is0.
In his paperRéflexions sur la résolution algébrique des équations ("Thoughts on the algebraic solving of equations"),[36]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 andquartic equations, but Lagrange did not succeed in applying it to aquintic equation, because it requires solving a resolvent polynomial of degree at least six.[37][38][39] 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 theAbel–Ruffini theorem. Nevertheless, modern methods for solving solvable quintic equations are mainly based on Lagrange's method.[39]
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 equationax3 +bx2 +cx +d = 0, but the computation is simpler with the depressed cubic equation,t3 +pt +q = 0.
Lagrange's main idea was to work with thediscrete Fourier transform of the roots instead of with the roots themselves. More precisely, letξ be aprimitive third root of unity, that is a number such thatξ3 = 1 andξ2 +ξ + 1 = 0 (when working in the space ofcomplex numbers, one has but this complex interpretation is not used here). Denotingx0,x1 andx2 the three roots of the cubic equation to be solved, letbe the discrete Fourier transform of the roots. Ifs0,s1 ands2 are known, the roots may be recovered from them with the inverse Fourier transform consisting of inverting this linear transformation; that is,
ByVieta's formulas,s0 is known to be zero in the case of a depressed cubic, and−b/a for the general cubic. So, onlys1 ands2 need to be computed. They are notsymmetric functions of the roots (exchangingx1 andx2 exchanges alsos1 ands2), but some simple symmetric functions ofs1 ands2 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 thesi 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 thesi 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, and are such symmetric polynomials (see below). It follows that and are the two roots of the quadratic equation Thus the resolution of the equation may be finished exactly as with Cardano's method, with and in place ofu and
In the case of the depressed cubic, one has and while in Cardano's method we have set and Thus, up to the exchange ofu and we have and 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.
A straightforward computation using the relationsξ3 = 1 andξ2 +ξ + 1 = 0 givesThis shows thatP andS are symmetric functions of the roots. UsingNewton's identities, it is straightforward to express them in terms of theelementary symmetric functions of the roots, givingwithe1 = 0,e2 =p ande3 = −q in the case of a depressed cubic, ande1 = −b/a,e2 =c/a ande3 = −d/a, in the general case.
Marden's theorem states that thefoci of theSteiner inellipse of any triangle can be found by using the cubic function whose roots are the coordinates in thecomplex plane of the triangle's three vertices. The roots of thefirst derivative of this cubic are the complex coordinates of those foci.
Thearea of a regularheptagon can be expressed in terms of the roots of a cubic. Further, the ratios of the long diagonal to the side, the side to the short diagonal, and the negative of the short diagonal to the long diagonal all satisfy a particular cubic equation. In addition, the ratio of theinradius to thecircumradius of aheptagonal triangle is one of the solutions of a cubic equation. The values of trigonometric functions of angles related to satisfy cubic equations.
Given the cosine (or other trigonometric function) of an arbitrary angle, the cosine ofone-third of that angle is one of the roots of a cubic.
^InO'Connor, John J.;Robertson, Edmund F.,"Omar Khayyam",MacTutor History of Mathematics Archive,University of St Andrews one may readThis problem in turn led Khayyam to solve the cubic equationx3 + 200x = 20x2 + 2000and he found a positive root of this cubic by considering the intersection of a rectangular hyperbola and a circle. An approximate numerical solution was then found by interpolation in trigonometric tables. The geometric construction was perfectly suitable for Omar Khayyam, as it occurs for solving a problem of geometric construction. At the end of his article he says only that, for this geometrical problem, if approximations are sufficient, then a simpler solution may be obtained by consultingtrigonometric tables. Textually:If the seeker is satisfied with an estimate, it is up to him to look into the table of chords of Almagest, or the table of sines and versed sines of Mothmed Observatory. This is followed by a short description of this alternate method (seven lines).
^More precisely, Vieta introduced a new variablew and imposed the conditionw(t +w) =p/3. This is equivalent with the substitutiont =p/3w −w, and differs from the substitution that is used here only by a change of sign ofw. This change of sign allows getting directly the formulas of§ Cardano's formula.
^Archimedes (October 8, 2007).The works of Archimedes. Translation by T. L. Heath. Rough Draft Printing.ISBN978-1603860512.
^Mikami, Yoshio (1974) [1913], "Chapter 8 Wang Hsiao-Tung and Cubic Equations",The Development of Mathematics in China and Japan (2nd ed.), New York: Chelsea Publishing Co., pp. 53–56,ISBN978-0-8284-0149-4
^A paper of Omar Khayyam, Scripta Math. 26 (1963), pages 323–337
^Guilbeau (1930, p. 9) states, "Omar Al Hay of Chorassan, about 1079 AD did most to elevate to a method the solution of the algebraic equations by intersecting conics."
^Berggren, J. L. (1990), "Innovation and Tradition in Sharaf al-Dīn al-Ṭūsī's Muʿādalāt",Journal of the American Oriental Society,110 (2):304–309,doi:10.2307/604533,JSTOR604533
^These are Formulas (80) and (83) of Weisstein, Eric W. 'Cubic Formula'. From MathWorld—A Wolfram Web Resource.https://mathworld.wolfram.com/CubicFormula.html, rewritten for having a coherent notation.
^Holmes, G. C., "The use of hyperbolic cosines in solving cubic polynomials",Mathematical Gazette 86. November 2002, 473–477.
^Abramowitz, Milton; Stegun, Irene A., eds.Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover (1965), chap. 22 p. 773
^Lagrange, Joseph-Louis (1869) [1771], "Réflexions sur la résolution algébrique des équations", inSerret, Joseph-Alfred (ed.),Œuvres de Lagrange, vol. III, Gauthier-Villars, pp. 205–421