Algebraic Solution of Cubics

A little history: Omar Khayyam (1048-1131, Persian, best known in the West for his philosophical poetry The Rubaiyat) published a method for geometrically constructing the roots of cubic equations. Most historical accounts assert correctly that Khayyam did not consider the negative roots of cubics. However, they are misleading in that they all fail to mention that his methods are fully sufficient to find the negative roots. For a description of Khayyam's methods see

http://www.math.cornell.edu/~dwh/papers/geomsolu/geomsolu.html.

This is in contrast to the common assertion that Girolamo Cardano (1501-1578, Italian) was the first to publish the general solution of cubic equations when in fact, as we shall see, he himself admitted that his methods are insufficient to find the real roots of many cubics.

Cardano published his algebraic solutions in his book, Artis Magnae (The Great Art) which was published in 1545. For a readable translation and historical summary see [Cardano (edited by Witmer), The Great Art, MIT Press, 1968]. Cardano used only positive coefficients and thus divided the cubic equations into 13 types (for example, he considered x³ + ax = b, x³ + b = ax, x³ = ax + b, and x³ + ax + b = 0 to be a different types). Cardano also used geometry to prove his solutions for each type. We allow ourselves the convenience of using negative numbers and lengths, and so have only one type of cubic: x³ + px² + qx + r = 0, where now we allow p, q, r to be either negative or positive.

a. Show that the substitution x = z - (p/3) reduces the general cubic equation to one of the form z³ + az + b = 0.

The main "trick" that Cardano used was to assume that there is a solution of z³ + az + b = 0 of the form z = t^1/3 + u^1/3. Plugging this into the cubic we get

(t^1/3 + u^1/3)³ + a(t^1/3 + u^1/3) + b = 0.

If you expand and simplify this, you get to

t + u + b + (3t^1/3u^1/3 + a)(t^1/3 + u^1/3) = 0.

(Cardano did this expansion and simplification geometrically by imagining a cube with sides t^1/3 + u^1/3.) Thus z = t^1/3 + u^1/3 is a root if

t + u = - b and t^1/3 u^1/3 = -(a/3).

b. Solving the above, show that t and u are the roots of the quadratic equation

w² + bw - (a/3)³ = 0.

Now solve this to get

and .

Thus the cubic has roots

z = t^1/3 + u^1/3

= { }^1/3 + { }^1/3.

This is Cardano's cubic formula. But, a strange thing happened. Cardano noticed that the cubic z³ = 15z + 4 has a positive real root 4 but, for this equation, a = -15 and b = -4, and if we put these values into his cubic formula, we get that the roots of z³ = 15z + 4 are

z  =  { }^1/3 + { }^1/3 .

How can this expression yield 4? In fact, this expresses the roots as the sum of two complex numbers even though you can show:

c. All three roots of z³ = 15z + 4 are real. [Divide the equation by (z - 4) and solve the resulting quadratic by using the quadratic formula.]

In Cardano's time there was no theory of complex numbers and so he reasonably concluded that his method would not work for this equation, even though he did investigate expressions such as . Cardano writes ([Cardano, page 103]):

When the cube of one-third the coefficient of x is greater than the square of one-half the constant of the equation ... then the solution of this can be found by the aliza problem which is discussed in the book of geometrical problems.

It is not clear what book he is referring to, but the "aliza problem" presumably refers to the mathematician known as al'Hazen, Abu Ali al'Hasan ibu al'Haitam (965-1039), who was born in Persia and worked in Egypt and whose works were known in Europe in Cardano's time. Al'Hazen had used intersecting conics to solve specific cubic equations and the problem of describing the image seen in a spherical mirror  this later problem is in some books called "Alhazen's problem."

In addition, we know today that each complex number has three cube roots and so the formula

z = { }^1/3 + { }^1/3

is ambiguous. In fact, some choices for the two cube roots give roots of the cubic and some do not. (Experiment with z³ = 15z + 4.) Faced with Cardano's Formula and equations like z³ = 15z + 4, Cardano and other mathematicians of the time started exploring the possible meanings of these complex numbers and thus started the theory of complex numbers.

d. Solve the cubic z³ = 15z + 4 using Cardano's Formula and your knowledge of complex numbers.

Remember that on the previous page we showed that z = t^1/3 + u^1/3 is a root of the equation if t + u = -b and t^1/3 u^1/3 = -(a/3).