4.5: Cubic equations

The first recorded procedure for finding the positive roots of any given quadratic equation dates from around 1700 BC (ancient Babylonian). A corresponding procedure for cubic equations had to wait until the early sixteenth century AD. The story is a slightly complicated one - involving public contests, betrayal, and much else besides.

In Section 4.4 we saw that the cubic equation x³ = 1 has three solutions - two of which are complex numbers. But in the sixteenth century, even negative numbers were viewed with suspicion, and complex numbers were still unknown. Moreover, symbolical algebra had not yet been invented, so everything was carried out in words: constants were “numbers”; a given multiple of the unknown was referred to as so many “things”; a given multiple of the square of the unknown was simply referred to as “squares”; and so on.

In short, we know that an improved method for sometimes finding a (positive) unknown which satisfied a cubic equation was devised by Scipione del Ferro (1465–1526) around 1515. He kept his method secret until just before his death, when he told his student Antonio del Fiore (1506-??). Niccolo Tartaglia (1499–1557) then made some independent progress in solving cubic equations. At some stage (around 1535) Fiore challenged Tartaglia to a public “cubic solving contest”. In preparing for this event, Tartaglia managed to improve on his method, and he seems to have triumphed in the contest. Tartaglia naturally hesitated to divulge his method in order to preserve his superiority, but was later persuaded to communicate what he knew to Girolamo Cardano (1501–1576) after Cardano promised not to publish it (either never, or not before Tartaglia himself had done so). Cardano improved the method, and his student Ferrari (1522–1565) extended the idea to give a method for solving quartic equations - all of which Cardano then published, contrary to his promise, but with full attribution to the rightful discoverers, in his groundbreaking book Ars Magna (1545 -just two years after Copernicus (1473–1543) published his De revolutionibus ...). Problem 134 illustrates the necessary first move in solving any cubic equation. Problem 135 then illustrates the general method in a relatively simple case.

Problem 134

(a) Given the equation x³ + 3x² − 4 = 0, choose a constant a, and then change variable by substituting y = x + a to produce an equation of the form y³ + ky = constant.

(b) In general, given any cubic equation ax³ + bx² + c^x + d = 0 with $a\ne 0$ , show how to change variable so as to reduce this to a cubic equation with no quadratic term.

Problem 135 The equation x³ + 3x² − 4 = 0 clearly has “x = 1” as a positive solution. (The other two solutions are x = −2, and x = −2 - a repeated root; however negatives were viewed with suspicion in the sixteenth century, so this root might well have been ignored.) Try to understand how the following sequence of moves “finds the root x = 1”:

(i) substitute y = x + 1 to get a cubic equation in y with no term in y²;

(ii) imagine y = u + v and interpret the identity for

$${(u+v)}^3=u^3+3uv(u+v)+v^3$$

as your cubic equation in y;

(iii) solve the simultaneous equations “3uv = 3”, “u³ + v³ = 2” (not by guessing, but by substituting $v=\frac{1}{u}$ from the first equation into the second to get a quadratic equation in “u³”, which you can then solve for u³ before taking cube roots);

(iv) then find the corresponding value of v, hence the value of y = u + v, and hence the value of x.

The simple method underlying Problem 135 is in fact completely general. Given any cubic equation

$$a{x}^3+b{x}^2+cx+d=0(\mathrm{with}a\ne 0)$$

we can divide through by a to reduce this to

$$x^3+p{x}^2+qx+r=0$$

with leading coefficient = 1. Then we can substitute $y=x+\frac{p}{3}$ and reduce this to a cubic equation in y

$$y^3-3{\left(\frac{p}{3}\right)}^{2}y+qy+\left[r+2{\left(\frac{p}{3}\right)}^3-q\left(\frac{p}{3}\right)\right]=0$$

which we can treat as having the form

$$y^3-my-n=0.$$

So we can set y = u + v (for some unknown u and v yet to be chosen), and treat the last equation as an instance of the identity

$${(u+v)}^3-3uv(u+v)-\left(u^3+v^3\right)=0$$

which it will become if we simply choose u and v to solve the simultaneous equations

$$3uv=m,u^3+v^3=n.$$

We can then solve these equations to find u, then v - and hence find y = u + v and $x=y-\frac{p}{3}$ .