The Renaissance Solution of the Cubic Equation II

We begin with a short discussion of the binomial theroem. Consider Pascal's triangle



...1...
..1.1..
.1.2.1.
1.3.3.1




Each row in the triangle is obtained by summing adjacent elements in the row above it. The binomial theorem says that the coefficients for the expansion of (x + y)n are given by the corresponding elements in Pascals triangle. For instance,

(x + y)2 = x2 + 2 x y + y2,

has coefficients 1,2,1 as in row three, and

(x + y)3 = x3 + 3 x 2 y + 3x y2 + y3,

has coefficients 1,3,3,1 as in row four. Now let us use this fact to expand (x +1)3

(x +1)3 = x3 + 3 x 2 (1) + 3 x (1)2 + (1)3,

or

(x +1)3 = x3 + 3 x 2 + 3 x + 1.

Next we expand (x +5)3

(x +5)3 = x3 + 3 x 2 (5) + 3 x (5)2 + (5)3.

(x +5)3 = x3 + 15 x 2 + 75 x + 125.

Now we wish to study the equation

x3+ p x2= N.

This is Tartaglia's equation. We will show how Tartaglia solved this equation. The equation can be solved by writing it in a different form. We will convert this equation into the equation of Del Ferro. We begin with a change of variable. Let us assume that our new variable differs from x by a constant, or

x = z + k.

Then we have

(z + k)3 + p(z + k)2 = N.

By expanding the powers of z, we get

z3 + 3z2k + 3zk2 + k3 + pz2 + 2pzk + pk2 = N.

Gathering together like terms of powers of z, we have

z3 + (3k + p) z2 + (3k2 + 2pk) z = N-k3 - pk2.

I want the z2 term to disappear! What should I do? Let k=-p/3, and we have,

z3 + 0 z2 + (3p2/9 - 2p2/3) z = N + p3/27 - p3/9.

Simplifying this, we obtain

z3 - (p2 /3)z= N - 2 p3/27,

which is just Del Ferro's eqution with,

m = -p2/3

and

n = N - 2p3/27

A solution is given by computing,

a3 = N/2 + p3/27 + ( (N/2 - p3/27)2 - p6/729) 1/2

and

b3 = -N/2 + p3/27=( (N/2 - p3/27) 2 - p6/729) 1/2

If we can compute a and b from the equations above, then z = a- b is a solution of Del Ferro's equation, and x = a- b - (p/3) is a solution of Tartaglia's equation. Let's consider an example:

x3 + 9x2 = 54,

then p = 9, and N = 54. Accordingly, we should take k = -9/3, or k= -3. Therefore, we make the substitution:

x = z - 3

which gives

(z - 3)3 + 9 (z - 3)2 = 54.

Expanding using the binomial formula we obtain the following:

z3 + 3z2 (-3) + 3z(-3)2 - 27 + 9z2 - 9(6)z + 81 = 54,

and

z3 + 27z - 27 - 54z + 81 = 6

which simplifies as

z3 - 27z = 54 - 81 + 27

or

z3 - 27z = 0

This factors as

(z2 - 27)z = 0.

One solution is z =0 and the other solutions are the plus or minus the square root of 27.

We have nearly completed our study of the cubic equation. Finally, let us consider the most general form of the cubic, namely:

x3+ p x2 + Mx= N.

This equation can also be reduced to the form of an equation of Del Ferro simply by making the substitution

x = z - p/3

Once again, we will obtain the equation

z3 + (M - p2 /3)z= N + Mp/3 - 2 p3/27.

which can be solved using the method of Del Ferro.

Homework Assignment 4

  1. Convert the following to Del Ferro's Equation. Do not solve.

    1. x3 + 3x2 = 54

    2. x3 - 9x2 = 27

  2. Solve the cubic equations using the method of substitution..

    1. x3 + 6x2 = 16

    2. x3 +3x2 - 6x2 = 8

To continue see The Greek Approach to the Quadratic Equation