Cubic Equations

The cubic formula is less mysterious if one does not begin by trying to guess a root. Instead, one changes variables until the cubic has a form in which its nonlinear part can be split into two pieces. The key point is that all of Cardano’s formula, Viète’s substitution, the trigonometric solution, and Lagrange resolvents reduce the cubic to a quadratic equation.

We begin with the monic cubic

\displaystyle x^3+a x^2+b x+c=0.

The quadratic term is an inconvenience, but it can always be removed by translating the variable. Use shifts { x=y+t} to eliminate the {x^2} term. { (y+t)^{3}+a (y+t)^{2}+b (y+t)+c=0} doesn’t have {y^2} term if {3t+a =0.} Therefore {x=y-\frac{a}{3}.} Taking x=y-\frac a3, expanding and collecting terms gives a depressed cubic

\displaystyle y^3+py+q=0,

where p=b-\frac{a^2}{3}, q=c-\frac{ab}{3}+\frac{2a^3}{27}. Thus every cubic is reduced to understanding the equation y^3+py+q=0.

Cardano’s substitution

The expression y^3+py suggests writing y=u+v. Indeed,

\displaystyle (u+v)^3+p(u+v)+q =\big(u^3+v^3+q\big)+\big(3uv+p\big)(u+v).

This identity is the heart of Cardano’s method. We try to choose u and v so that the two brackets vanish separately: u^3+v^3=-q,\quad uv=-\frac p3. Now set U=u^3,\quad V=v^3. Then

\displaystyle U+V=-q,\quad UV=-\frac{p^3}{27}.

So U and V are the two roots of the quadratic equation

\displaystyle Z^2+qZ-\frac{p^3}{27}=0.

Therefore

\displaystyle U=-\frac q2+\sqrt{\big(\frac q2\big)^2+\big(\frac p3\big)^3},\quad V=-\frac q2-\sqrt{\big(\frac q2\big)^2+\big(\frac p3\big)^3}.

Introduce the discriminant D:=\left(\frac q2\right)^2+\left(\frac p3\right)^3.

Choose a cube root u of U. The corresponding cube root v must not be chosen independently: it must satisfy uv=-\frac p3. Thus, provided u\ne0, one should define v=-\frac{p}{3u}. Then v^3=V, and y=u+v is a root of the depressed cubic. Hence one root of the original equation is

\displaystyle x=-\frac a3+ \sqrt[3]{-\frac q2+\sqrt{D}} +\sqrt[3]{-\frac q2-\sqrt{D}}.

Let \omega=e^{2\pi i/3}. If u^3=U and v^3=V are chosen with uv=-p/3, then all three roots of the depressed cubic are

\displaystyle y_k=\omega^k u+\omega^{-k}v,\quad k=0,1,2.

Consequently the three roots of the original cubic are

\displaystyle x_k=-\frac a3+\omega^k u+\omega^{-k}v,\quad k=0,1,2.

The important qualification is that the two cube roots must be matched so that their product is -p/3. Arbitrary choices of the two cube roots generally do not solve the equation.

Viète’s substitution

Viète’s substitution is not a different miracle; it builds the condition uv=-p/3 directly into the notation. Write u=z, ~v=-p/3z to get

\displaystyle y=z-\frac{p}{3z},\quad z\ne0.

Then y^3+py =\big(z-\frac{p}{3z}\big)^3+p\big(z-\frac{p}{3z}\big) =z^3-\frac{p^3}{27z^3}. Thus the depressed cubic becomes

\displaystyle z^3-\frac{p^3}{27z^3}+q=0.

Multiplying by z^3 gives z^6+qz^3-\frac{p^3}{27}=0. This is quadratic in z^3. Hence

\displaystyle z^3=-\frac q2\pm\sqrt{\big(\frac q2\big)^2+\big(\frac p3\big)^3}.

After choosing a cube root z, one obtains a root

\displaystyle y=z-\frac{p}{3z}.

The relation with Cardano is immediate: set u=z,~ v=-\frac{p}{3z}. Then y=u+v and uv=-p/3 automatically. Viète’s substitution is therefore Cardano’s substitution with the compatibility condition already enforced. It might look like there are two choice for z^3 from the quadratic roots. But choice of the square root merely exchange the roles of the two Cardano quantities u and v. and doesn’t produce any new quantities.

The trigonometric solution

Suppose now that p<0. Write y=2\sqrt{-\frac p3}\cos\theta. Using \cos(3\theta)=4\cos^3\theta-3\cos\theta, one finds

\displaystyle y^3+py =2\big(-\frac p3\big)^{3/2}\cos(3\theta).

Therefore y^3+py+q=0 is equivalent to

\displaystyle \cos(3\theta) =\frac{3q}{2p}\sqrt{-\frac3p}.

When \big|\frac{3q}{2p}\sqrt{-\frac3p}\big|\le1, there are three real solutions. If \theta_0=\frac13\arccos\big(\frac{3q}{2p}\sqrt{-\frac3p}\big), then the three roots of the depressed cubic are

\displaystyle y_k= 2\sqrt{-\frac p3}\cos\big(\theta_0+\frac{2\pi k}{3}\big), \quad k=0,1,2.

This is especially useful in the case where all three roots are real. Cardano’s formula still works there, but the square root \sqrt{D} is imaginary when D<0, even though the final answers are real. This phenomenon is the classical casus irreducibilis: real roots are naturally described by trigonometric functions, while the radical formula must pass through complex numbers.

At first this looks useful only when p<0, since then s=\sqrt{-p/3} is real. But algebraically the substitution works whenever p\ne0: if p>0, then s is imaginary, and if the cubic has non-real roots, then the angle \theta is simply allowed to be complex. Over \mathbb C, the equation \cos W=C is solved exactly as one solves a quadratic. Put z=e^{iW}. Then \cos W=\frac{z+z^{-1}}2=C, so z^2-2Cz+1=0. Hence z=C\pm\sqrt{C^2-1}. Choose either root \lambda of this quadratic and choose a logarithm such that e^{iW_0}=\lambda. Equivalently, W_0=-i\log\left(C+\sqrt{C^2-1}\right). Then

\displaystyle \theta_k=\frac{W_0+2\pi k}{3},\quad y_k=2\sqrt{-\frac p3}\cos\theta_k,\qquad k=0,1,2,

are the three roots. Choosing the other square-root branch replaces W_0 by -W_0 modulo 2\pi; this merely permutes the same three values of y_k. Likewise, changing the branch of the logarithm adds a multiple of 2\pi to W_0, which again only permutes the three roots.

Put z=e^{i\theta}. Then y=2\sqrt{-\frac p3} \cos\theta=\sqrt{-\frac p3}\left(z+\frac1z\right). So the substitution gives y^3+py=s^3\left(z^3+\frac1{z^3}\right). Thus the cubic is equivalent to z^3+\frac1{z^3}=-\frac q{s^3}, or z^6+\frac q{s^3}z^3+1=0. So the trigonometric method is again a quadratic equation in z^3. This is the same mechanism as Cardano and Viète, written in the multiplicative variable z=e^{i\theta}.

Lagrange resolvents

Cardano’s computation can also be understood from the roots themselves. Let x_1,x_2,x_3 be the roots of x^3+a x^2+b x+c=0. Define the Fourier-type combinations

\displaystyle R_0=x_1+x_2+x_3,

\displaystyle R_1=x_1+\omega x_2+\omega^2x_3,

\displaystyle R_2=x_1+\omega^2x_2+\omega x_3.

The first of these is the familiar symmetric quantity R_0=x_1+x_2+x_3=-a. The inverse transform recovers the roots:

\displaystyle x_1=\frac{R_0+R_1+R_2}{3},

\displaystyle x_2=\frac{R_0+\omega^2R_1+\omega R_2}{3},

\displaystyle x_3=\frac{R_0+\omega R_1+\omega^2R_2}{3}.

Thus it suffices to understand R_1 and R_2. Although these quantities themselves depend on an ordering of the roots, the expressions R_1R_2 and R_1^3+R_2^3 are symmetric. A direct calculation gives

\displaystyle R_1R_2=x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1 =a^2-3b,

\displaystyle R_1^3+R_2^3 =2(x_1+x_2+x_3)^3-9(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+27x_1x_2x_3,

so that R_1^3+R_2^3=-2a^3+9ab-27c. Also, R_1^3R_2^3=(R_1R_2)^3=(a^2-3b)^3. Therefore R_1^3 and R_2^3 are the roots of the quadratic equation

\displaystyle Z^2-\left(-2a^3+9ab-27c\right)Z+(a^2-3b)^3=0.

This is the resolvent quadratic. Once it is solved, one chooses compatible cube roots R_1,R_2, uses the inverse transform above, and obtains x_1,x_2,x_3.

After translating to the depressed cubic, this resolvent quadratic becomes exactly Cardano’s quadratic. Thus Cardano’s substitution is not an isolated trick: it is the root-level statement that the nontrivial transforms of the three roots can be cubed to produce symmetric quantities.

Discriminant and the geometry of the roots

For a general cubic f(x)=Ax^3+Bx^2+Cx+D,\quad A\ne0, the discriminant is

\displaystyle \Delta= 18ABCD-4B^3D+B^2C^2-4AC^3-27A^2D^2.

Equivalently, \Delta=A^4(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2. Thus \Delta=0 precisely when two roots coincide. For a cubic with real coefficients: \Delta>0 means that there are three distinct real roots; \Delta<0 means that there is one real root and one non-real conjugate pair; \Delta=0 means that the cubic has a repeated root.

For the depressed cubic y^3+py+q, the discriminant is \Delta=-4p^3-27q^2=-108 D. Thus Cardano’s quantity D is simply the discriminant in a different normalization. The three cases \mathcal D<0, \mathcal D>0, and \mathcal D=0 correspond respectively to three distinct real roots, one real root, and repeated roots.

The derivative also gives a useful geometric picture: f'(x)=3Ax^2+2Bx+C. Its discriminant is 4(B^2-3AC). Hence B^2-3AC>0 means that the graph has two distinct critical points; B^2-3AC<0 means that the graph is strictly monotone and therefore has exactly one real root; B^2-3AC=0 means that the two critical points merge into a stationary point of inflection.

When \Delta=0, the repeated-root case separates further. If B^2-3AC>0, there is one double root and one distinct simple root; if B^2-3AC=0, all three roots coincide.

TARTAGLIA’S ORIGINAL POEM

Quando chel cubo con le cose appresso
Se aqquaglia ?a qualche numero discreto
Trouan duo altri differenti in esso
Dapoi terrai questo per consueto
Che?llor productto sempre sia equale
Alterzo cubo delle cose neto,
El residuo poi suo generale
Delli lor lati cubi ben sottrati
Varra la tua cosa principale.
In el secondo de cotestiatti
Quando chel cubo restasse lui solo
Tu osseruarai questaltri contratti,
Del numer farai due tal part?`a uolo
Che luna in laltra si produca schietto
El terzo cubo delle cose in stolo
Delle qual poi, per communprecetto
Torrai li lati cubi insieme gionti
Et cotal somma sara il tuo concetto.
El terzo poi de questi nostri conti
Se solue col secondo se ben guardi
Che per natura son quasi congionti.
Questi trouai, non con passi tardi
Nel mille cinquecent`e, quatroe trenta
Con fondamenti ben sald`e gagliardi
Nella citta dal marintorno centa.

It translates to: When the cube together with the linear term equals a number, find two different numbers whose product is the cube of one third of the linear coefficient; subtract their cube roots, and you obtain the unknown. When the cube stands alone, divide the number into two parts with the same prescribed product; add their cube roots, and this sum is the answer. The third case is solved like the second, since the two are closely related. I discovered these rules in 1534, in Venice.

So Tartaglia’s famous verse describes, in sixteenth-century language, the essential Cardano step. The three cases arise because algebra generally treated the possible placements of positive terms separately. In modern symbols they are \displaystyle x^3+Ax=B, ~ x^3=Ax+B, ~ x^3+B=Ax. In the first case, Tartaglia writes x=u-v and seeks U=u^3, V=v^3 satisfying U-V=B,\qquad UV=(A/3)^3; then x=\sqrt[3]{U}-\sqrt[3]{V}. In the second case he writes x=u+v, so that U+V=B,\qquad UV=(A/3)^3, and x=\sqrt[3]{U}+\sqrt[3]{V}. The third case reduces to the second after replacing x by -x. The reason behind all three rules is the expansion (u\pm v)^3=u^3\pm v^3+3uv(u\pm v): imposing \displaystyle uv=A/3 makes the last term exactly the required linear term.

One seeks two quantities whose sum is determined by the constant term and whose product is determined by the coefficient of the linear term. In modern notation, one solves

\displaystyle U+V=-q,\quad UV=-\frac{p^3}{27},

takes cube roots u^3=U, v^3=V, with uv=-p/3, and then obtains the desired quantity as y=u+v. The historical formula is therefore already the modern one in compressed form: the cubic is solved by carefully choosing a decomposition converts it into a quadratic equation.

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