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# Cubic equation

A cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. An example is the equation

2x3 - 4x2 + 3x - 4 = 0

and the general form may be written as

α3x3 + α2x2 + α1x + α0 = 0.

Usually, the coefficients α0, ..., α3 are real numbers. However, most of the theory is also valid if they belong to a field of characteristic other than two or three. We will always assume that α3 is non-zero (otherwise it is a quadratic equation).

Solving a cubic equation amounts to finding the roots of a cubic function.

 Contents

## History

The Persian mathematician Omar Khayyám (1048–1123) constructed solutions of cubic equations by intersecting a conic section with a circle. He showed how this geometric solution could be used to get a numerical answer by consulting trigonometric tables.

In the early sixteenth century, the Italian mathematician Scipione dal Ferro found a method for solving a class of cubic equations, namely those x3 + mx = n. In fact, all cubic equations can be reduced to this form if we allow m and n to be negative, but negative numbers were not known at that time. Dal Ferro kept his achievement secret until just before his death, when he told his student about it. Tartaglia heard about this and soon found a method himself. He revealed his method to Gerolamo Cardano, who published it in Ars Magna 1545.

Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with these complex numbers in Ars Magna, but he did not really understand it. Rafael Bombelli studied this issue in detail and is therefore often considered as the discoverer of complex numbers.

## The nature of the roots

Every cubic equation with real coefficients has at least one solution x among the real numbers. We can distinguish the different possible cases in of the discriminant

$\Delta = 4\alpha_1^3\alpha_3 - \alpha_1^2\alpha_2^2 + 4\alpha_0\alpha_2^3 - 18\alpha_0\alpha_1\alpha_2\alpha_3 + 27\alpha_0^2\alpha_3^2.$

The following cases need to be considered.

• If Δ < 0, then the equation has three distinct real roots.
• If Δ > 0, then the equation has one real roots and a pair of complex conjugate roots.
• If Δ = 0, then (at least) two roots coincide. We now define
$\Delta_2 = 2\alpha_2^3 - 9\alpha_1\alpha_2\alpha_3 + 27\alpha_0\alpha_3^2.$
If Δ2 vanishes as well, then all three roots coincide and we have a triple real root. Otherwise, the equation has a double real root and a single real root. The number Δ2 is the resultant of the cubic and its second derivative.

## Cardano's method

The solutions can be found with the following method due to Scipione dal Ferro and Tartaglia, published by Gerolamo Cardano in 1545.

We first divide the given equation by α0 to arrive at an equation of the form

x3 + ax2 + bx + c = 0.         (1)

The substitution x = t - a/3 eliminates the quadratic term; in fact, we get the equation

$t^3 + pt + q = 0, \quad\mbox{where } p = b - \frac{a^2}3 \quad\mbox{and}\quad q = c + \frac{2a^3-9ab}{27}. \qquad (2)$

This is called the depressed cubic.

Suppose that we can find numbers u and v such that

$u^3-v^3 = q \quad\mbox{and}\quad uv = \frac13p. \qquad (3)$

A solution to our equation is then given by

t = vu,

as can be checked by directly substituting this value for t in (2).

The system (3) can be solved by solving the second equation for v, which gives

$v = \frac{p}{3u}.$

Substituting this in the first equation in (3) yields

$u^3 - \frac{p^3}{27u^3} = q.$

This can be seen as a quadratic equation for u3. If we solve this equation, we find that

$u=\sqrt[3]{{q\over 2}\pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}. \qquad (4)$

Since t = vu and t = x + a/3, we find

$x=\frac{p}{3u}-u-{a\over 3}.$

Note that there are six possibilities in computing u with (4), since there are two solutions to the square root, and three complex solutions to the cubic root. However, which solution to the square root is chosen does not affect the final resulting x.

## Lagrange resolvents

The symmetric group S3 of order three has the cyclic group of order three as a normal subgroup, which suggests making use of the discrete Fourier transform of the roots, an idea due to Lagrange. Suppose that r0, r1 and r2 are the roots of equation (1), and define $\zeta = (1+\sqrt{3}i)/2$, so that ζ is a primitive third root of unity. We now set

$s_0 = r_0 + r_1 + r_2,\,$
$s_1 = r_0 + \zeta r_1 + \zeta^2 r_2,\,$
$s_2 = r_0 + \zeta^2 r_1 + \zeta r_2.\,$

The roots may then be recovered from the three si by inverting the above linear transformation, giving

$r_0 = (s_0 + s_1 + s_2)/3,\,$
$r_1 = (s_0 + \zeta^2 s_1 + \zeta s_2)/3,\,$
$r_2 = (s_0 + \zeta s_1 + \zeta^2 s_2)/3.\,$

We already know the value s0 = −a, so we only need to seek values for the other two. However, if we take the cubes, a cyclic permutation leaves the cubes invariant, and a transposition of two roots exchanges s13 and s23, hence the polynomial

$(z-s_1^3)(z-s_2^3) \qquad (5)$

is invariant under permutations of the roots, and so has coefficients expressible in terms of (1). Using calculations involving symmetric functions or alternatively field extensions, we can calculate (5) to be

${z}^{2}+ \left( -9\,ba+2\,{a}^{3}+27\,c \right) z+ \left( {a}^{2}-3\,b\right)^{3}.$

The roots of this quadratic equation are

$\frac92\,ab-{a}^{3}- \frac{27}{2}\,c \pm \frac32\,\sqrt{3\Delta},$

where Δ is the discriminant defined above. Taking cube roots give us s1 and s2, from which we can recover the roots ri of (1).

## Factorization

If r is any root of (1), then we may factor using r to obtain

(x - r)(x2 + (a + r)x + b + ar + r2) = x3 + ax2 + bx + c.

Hence if we know one root we can find the other two by solving a quadratic equation, giving

$\frac12 \left(-a-r \pm \sqrt{-3r^2-2ar+a^2-4b}\right)$

for the other two roots. If we are finding the roots of a polynomial with real coefficients and one real root, we can find the real root purely in terms of the real (rather than complex) cube root function, or alternatively stated we can find the root by extracting cube roots only of positive quantities. The complex conjugate roots can then be found as above.

The cube root function is in some respects not a well-behaved function, or one convenient for the purposes of finding the roots of a cubic equation. While cube roots are well-known and traditional, it is possible to use other algebraic functions to determine the roots, and avoid some of the problems of cube roots. The cube root function has a branch singularity at zero, as a result of which the real cube root function does not extend nicely to a complex cube root function. Moreover, when using cube roots to find the roots of a polynomial with three real roots we must take the roots of complex numbers, which introduces complex numbers into a situation which does not, in fact, require them.

We can get around these problems by using Chebyshev cube roots in place of ordinary cube roots. The polynomial C3 = x3 - 3x is the third Chebyshev polynomial normalized to the interval [-2, 2], which we use in place of [-1, 1] so as to obtain a monic polynomial. This polynomial function is equal to

$C_3(x) = t = 2 \cosh(\operatorname{arccosh}({x\over2})/3).$

Inverting this function, we obtain

$x = 2 \operatorname{arccosh}(\cosh({t\over2})/3),$

which satisfies x3 - 3x = t.

This procedure is precisely analogous to the definition of the cube root in terms of logarithms and exponentials, with 2 arccosh(x) in the place of ln(x), and cosh(x/2) in the place of exp(x). However, it tangles us up in the question of the branch cuts of arccosh. We can get around that by defining the Chebychev cube root in terms of the hypergeometric function

$C_{1\over3}(t) = F(-{1\over3}, {1\over3}, {1\over2}, \frac{2-x}{4})$

Rather than dealing with the complexities of either method, we can define the Chebyshev cube root for all complex values directly. For | t - 2 | < 4, we have the following convergent series:

$C_{1\over3}(t) = \sum_{n=0}^\infty \frac{2}{1-3n} {3n \choose n}(\frac{2-t}{27})^n.$

We can now define $C_{1\over3}(t)$ for the whole complex plane by taking a branch cut from $-\infty$ to -2, and analytically continuing; continuing to the branch cut itself through the upper half-plane.

Now if we want the roots of x3 - 3x - t we may find them in terms of $C_{1\over3}$ as $r_1 = C_{1\over3}(t),\,$ $r_2 = -C_{1\over3}(-t),\,$ r3 = - r1 - r2.

If we have a cubic equation in depressed form, we may write it as x3 - 3px + q = 0. Substituting $x = \sqrt{p} z$ we obtain $z^3 - 3z + p^{-\frac{3}{2}}q = 0$. From this we obtain solutions to our original equation in terms of the Chebyshev cube root as $r_1 = \sqrt{p}\,C_{1\over3}(p^{-\frac{3}{2}}q),\,$ $r_2 = -\sqrt{p}\,C_{1\over3}(-p^{-\frac{3}{2}}q),\,$ r3 = - r1 - r2. If now we start from equation (1) and reduce to the depressed form, we have p = (a2 - 3b) / 9 and q = (2a3 - 9ab + 27c) / 27, leading to

$t = p^{-\frac{3}{2}}q = \frac{2a^3-9ab+27c}{(a^2-3b)^{3/2}}.$

This gives us solutions to (1) as

$r_1 = \sqrt{p}\,C_{1\over3}(t),\,$
$r_2 = -\sqrt{p}\,C_{1\over3}(-t),\,$
r3 = - r1 - r2.

Suppose the coefficients of (1) are real. If s is the quantity q/r from the section on real roots, then s = t2; hence 0 < s < 4 is equivalent to -2 < t < 2, and in this case we have a polynomial with three distinct real roots, expressed in terms of a real function of a real variable, quite unlike the situation when using cube roots. If s > 4 then either t > 2 and $C_{1\over3}(t)$ is the sole real root, or t < -2 and $-C_{1\over3}(-t)$ is the role real root. If s < 0 then the reduction to Chebyshev polynomial form has given a t which is a pure imaginary number; in this case $iC_{1\over3}(-it)-iC_{1\over3}(it)$ is the sole real root. We are now evaluating a real root by means of a function of a purely imaginary argument; however we can avoid this by using the function

$S_{1\over3}(t) = iC_{1\over3}(-it)-iC_{1\over3}(it) = 2 \operatorname{arcsinh}(\sinh({t\over2})/3),\,$

which is a real function of a real variable with no singularities along the real axis. If a polynomial can be reduced to the form x3 + 3x - t with real t, this is a convenient way to solve for its roots.