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This article discusses cubic equations in one variable. For a discussion of cubic equations in two variables, see elliptic curve.

Graph of a cubic function; the roots are where the curve crosses the x-axis (y = 0). It has 2 critical points.

In mathematics, a cubic function is a function of the form

$f(x)=ax^3+bx^2+cx+d,,$

where a is nonzero; or in other words, a polynomial of degree three. The derivative of a cubic function is a quadratic function. The integral of a cubic function is a quartic function.

If you set $f (x) = 0$ , you get a cubic equation of the form:

$ax^3+bx^2+cx+d=0 ,$

where

$a e 0 ,$

(if a = 0, then the equation becomes a quadratic equation).

Usually, the coefficients $a$ , $b$ , $c$ , $d$ are real numbers. However, most of the theory is also valid if they belong to a field of characteristic other than two or three.

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

Roots of a cubic function

The nature of the roots

Every cubic equation with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant,

$Delta = 4b^3d - b^2c^2 + 4ac^3 - 18abcd + 27a^2d^2. ,$

The following cases need to be considered.

If ? < 0, then the equation has three distinct real roots.
If ? > 0, then the equation has one real root and a pair of complex conjugate roots.
If ? = 0, then (at least) two roots coincide. It may be that the equation has a double real root and another distinct single real root; alternatively, all three roots coincide yielding a triple real root. A possible way to decide between these subcases is to compute the resultant of the cubic and its second derivative: a triple root exists if and only if this resultant vanishes.

See also: multiplicity of a root of a polynomial

Cardano's method

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

We first divide the standard equation by the leading coefficient to arrive at an equation of the form

$x^3 + ax^2 + bx +c = 0. qquad (1)$

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

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

This is called the depressed cubic. A simple and elegant way of solving the depressed cubic is due to Thomas Harriot (1560 ? 1621): substituting $t=y-{pover 3y}$ into it and multiplying both sides by $y 3$ yields, after much cancellation, $y^6+q y^3-{p^3over 27}=0$ . Described below is the original, somewhat long-winded method of Cardano and Tartaglia, which still dominates the textbooks today.

Suppose that we can find numbers u and v such that

$u^3-v^3 = q quadmbox{and}quad uv = rac{p}{3}. quad (3)$

A solution to our equation is then given by

$t = v - u, ,$

as can be checked by directly substituting this value for t in (2), as a consequence of the third order binomial identity

$(v-u)^3+3uv(v-u)+(u^3-v^3)=0 .$

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

$v = rac{p}{3u}.$

Substituting this into the first equation in (3) yields

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

Moving the q to the other side and multiplying by 27u³ yields

$27u^6 - 27qu^3 - p^3 = 0,.$

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

$u^{3}={qover 2}pm sqrt{{q^{2}over 4}+{p^{3}over 27}}$

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

Since t = v ? u, t = x + a/3, and v = p/3u, we find

$x=- rac{p}{3u}+u-{aover 3}.$

Note that there are six possibilities in computing u with (4), since there are two solutions to the square root ( $pm$ ), and three complex solutions to the cubic root ? the principal root and the principal root multiplied by $frac{-1}{2} pm i frac{sqrt{3}}{2}$ . However, the sign of the square root (plus or minus) does not affect the final resulting x, although care must be taken in two special cases to avoid divisions by zero. First, if p = 0, then one should choose the positive square root so that u does not equal zero, i.e., $u = +sqrt[3]{q}$ . Second, if p = q = 0, then we have the triple real root x = ?a/3.

In summary, for the cubic equation

$x^3 + ax^2 + bx +c = 0$

the solutions for x are given by

$x=- rac{p}{3u}+u-{aover 3}$

where

$p = b - rac{a^2}3$

$q = c + rac{2a^3-9ab}{27}$

$u=sqrt[3]{-{qover 2}pm sqrt{{q^{2}over 4}+{p^{3}over 27}}}.$

Although this method is simple and elegant, it fails for the case of three real roots, e.g. when:

$D < 0, D = left ({q over 2} ight )^2 + left ({p over 3} ight )^3$

For this case a different method (e.g. goniometrical) has to be used.

Lagrange resolvents

The symmetric group S₃ 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 $zeta = (-1+isqrt{3})/2$ , so that ? 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 s_i 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

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

The roots of this quadratic equation are

$rac92,ab-{a}^{3}- rac{27}{2},c pm rac{1}{2},sqrt{D},$

where D is the discriminant. Taking cube roots give us s₁ and s₂, from which we can recover the roots r_i of (1).

Factorization

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

$left (x-r ight )left (x^2+(a+r)x+b+ar+r^2 ight ) = x^3+ax^2+bx+c.$

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

$rac12 left(-a-r pm sqrt{-3r^2-2ar+a^2-4b} ight)$

for the other two roots.

Root-finding formula

The formula for finding the roots of a cubic function is fairly complicated. Therefore, it is common to use the rational root test or a numerical solution instead.

If we have

$f(x) = ax^3 + bx^2 + cx + d = a(x - x_1)(x - x_2)(x - x_3),,$

let

$q = rac{3ac-b^2}{9a^2}$

and

$r = rac{9abc - 27a^2d - 2b^3}{54a^3}.$

Now, let

$s = sqrt[3]{r + sqrt{q^3+r^2}}$

and

$t = sqrt[3]{r - sqrt{q^3+r^2}}.$

The solutions are

$x_1 = s+t- rac{b}{3a},$

$x_2=-$

$x_3=-$

$q 3 + r 2$ is the discriminant, if it is less than 0 then all three solutions are real and unequal and the roots can then be expressed by trigonometric functions (trigonometric cosine).

Solution in terms of Chebyshev radicals

If we have a cubic equation which is already in depressed form, we may write it as $,x^3 - 3px - q = 0$ . Substituting $x = sqrt{p} z$ we obtain $z^3 - 3z - p^{- rac{3}{2}}q = 0$ or equivalently

$z^3 - 3z = p^{- rac{3}{2}}q .$

From this we obtain solutions to our original equation in terms of the Chebyshev cube root $C_{1over3}$ as

$r_0 = sqrt{p},C_{1over3}(p^{- rac{3}{2}}q),,$

$r_1 = -sqrt{p},C_{1over3}(-p^{- rac{3}{2}}q),,$

$r_2 = -r_0 - r_1 .$

If now we start from a general equation

$x^3 + ax^2 + bx +c = 0 qquad (1)$

and reduce it to the depressed form under the substitution x = t ? a/3, we have $, p = (a^2-3b)/9$ and $, q = -(2a^3-9ab+27c)/27$ , leading to

$t_{a;b;c} = p^{- rac{3}{2}}q = - rac{2a^3-9ab+27c}{(a^2-3b)^{3/2}}.$

This gives us the solutions to (1) as

$r_0 = sqrt{p},C_{1over3}(t_{a;b;c})-{aover 3} ,,$

$r_1 = -sqrt{p},C_{1over3}(-t_{a;b;c})-{aover 3},,$

$r_2 = -r_0 - r_1 - a .$

The case of a cubic equation with real coefficients

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 $C_{1over3}(t)$ is the sole real root, or t < ?2 and $-C_{1over3}(-t)$ is the sole 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 $iC_{1over3}(-it)-iC_{1over3}(it)$ 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

$S_{1over3}(t) = iC_{1over3}(-it)-iC_{1over3}(it) = 2 operatorname{sinh}left(operatorname{arcsinh}left({tover2} ight)/3 ight),,$

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 $x 3 + 3 x ? t$ with real t, this is a convenient way to solve for its roots.

Derivative

The derivative $f'(x)=3ax^2+2bx+c,$ will yield $x= rac{-b pm sqrt {b^2-3ac }}{3a}$ when $f'(x)=0,$ . Bearing its resemblance to the quadratic formula, this formula can be used to find the critical points of a cubic function. It turns out that, if $b^2-3ac > 0,$ , then the cubic function will have two critical points ? a local maximum and a local minimum; if $b^2-3ac = 0,$ , then there is one critical point, and it will yield the inflection point; and if $b^2-3ac < 0,$ , then there are no critical points.