
By §8, any number has three cube roots, two of which are the products
of the remaining one by the imaginary cube roots of unity:

(8)  \omega = -1/2 + 1/2 $$3i, \omega^2 = -1/2 - 1/2 $$3i.

We can choose particular cube roots

(9) A = 3$$-q/2 + $$R,  B = 3$$-q/2 - $$R,

such that AB = -p/3, since the product of the numbers under the cube
root radicals is equal to (-p/3)^3. Hence the six values of z are

A, \omega A, \omega^2 A,  B, \omega B, \omega^2 B.

These can be paired so that the product of the two in each pair is -p/3.

AB = -p/3,  \omega A · \omega^2 B = -p/3, \omega^2 A · \omega B = -p/3.

Hence with any root z is paired a root equal to -p/(3z). By (5), the sum
of the two is a value of y. Hence the three values of y are

(10) y_1 = A + B,  y_2 = \omega A + \omega^2 B, y_3 = \omega^2 A + \omega B.

It is easy to verify that these numbers are actually roots of (2). For
example, since \omega^3 = 1, the cube of y_2 is

A^3 + B^3 + 3\omega A^2 B + 3\omega^2 AB^2= -q - p(\omega A + \omega^2 B)= -q - py_2,

by (9) and AB = -p/3.

The numbers (10) are known as Cardan's formulas for the roots of a
reduced cubic equation (2). The expression A + B for a root was first
published by Cardan in his Ars Magna of 1545, although he had obtained
it from Tartaglia under promise of secrecy.

EXAMPLE.    Solve y^3 - 15y - 126 = 0.

Solution.   The substitution (5) is here y = z + 5/z.  We get

z^6 - 126z^3 + 125 = 0,  z^3 = 1 or 125.

The pairs of values of z whose product is 5 are 1 and 5, \omega and 5\omega^2, \omega^2 and 5\omega. Their
sums 6, \omega + 5\omega^2, and \omega^2 + 5\omega give the three roots.

EXERCISES

Solve the equations:

1. y^3 - 18y + 35 = 0.

2. x^3 + 6x^2 + 3x + 13[**unclear] = 0.

3. y^3 - 2y + 4 = 0.

4. 28x^3 + 9x^2 - 1 = 0.
