Cubic Equations

The iterative techniques used to solve quadratic equations allow a direct generalization to

ax³ + bx² + cx + d = 0.

We begin by extending the Babylonian technique for solving x² - k = 0 to x³ - k = 0. That is, if someone guesses "the cube root of 2 is 3/2," we check this initial guess x₁ = 3/2 and find that

(3/2)(3/2)(3/2) > 2.

While 3/2 is too large, there does exist a number for which

(3/2)(3/2)([number]) = 2,

and this number is 2/(x₁)² = 8/9. To arrive at a (hopefully improved) guess for x₂, we again seek a number between 8/9 and 3/2 - i.e., between x₁ and 2/(x₁)².

Here two possibilities present themselves. We could again choose x₂ as the arithmetic mean of x₁ and 2/(x₁)², arriving at

Alternatively, as suggested by the relation

we could choose x₂ as the arithmetic mean of the three numbers whose product is 2, arriving at

Generalizing from 2 to k, we are again led to two iterators,

both of which have k^1/3 as a fixed point.

While one can use calculus or spreadsheets to decide which of these iterators is "flatter" at the fixed point k^1/3, there is an algebraic alternative. For positive numbers a, b, and c, the Arithmetic-Geometric Mean Inequality asserts that

Exercise. Use the above inequality to show that F₂(x) has a minimum at x = k^1/3. [Hint: Set a = b = x and c = k/x².]

This technique for solving x³ - k = 0 by iteration extends readily to ax³ + d = 0 (set k = -d/a) and leads to the iterator

Again, we are faced with the problem of modifying this iterator to obtain one whose fixed points correspond to the solutions of

ax³ + bx² + cx + d = 0.

By using a spreadsheet to test various modifications, we are led to

as an iterator that appears to be flat at its fixed points for arbitrary values of a, b, c, and d.

It is on this basis that we are able to substitute the iterative scheme

for Cardano's method.

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