HEADERS_END

The next stage up from the quadratic equation, the general cubic

equation has the form EQN:ax^3+bx^2+cx+d=0.

Just as there is a solution to the quadratic equation, a general

solution to the cubic was found by Tartaglia and Ferro.

The problem is, even when there are three real solutions, the

intermediate calculations use the roots of negative numbers,

and this led directly to the acceptance of the complex numbers.

In practice, numerical solutions are found rather than using a

closed form formula. Newton's Method is an efficient way to find

solutions, although one must be aware that the basins of attraction

have fractal boundaries.

Also related: quartic equation

----

!! Outline of the solution

Start with a general cubic. Since the leading coefficient can't

be zero we can divide through by that, so our general cubic is:

* EQN:x^3+ax^2+bx+c=0

Substitute EQN:x=t-a/3 giving

* EQN:(t-a/3)^3+a(t-a/3)^2+b(t-a/3)+c=0

* EQN:t^3+pt+q=0

for suitable /p/ and /q./ Notice that there is no quadratic term.

Now substitute EQN:t=u-\frac{p}{3u} and multiply by EQN:u^3 to get

* EQN:u^6+qu^3+\frac{p^3}{27}=0

Now we have a quadratic in EQN:u^3 so using the quadratic equation

formula we get two possible answers for EQN:u^3. Each of these

gives three cube roots, giving six possible answers. These occur

in pairs of identical solutions, giving three in all as required.

The Wikipedia and MathWorld articles have more information.

* http://en.wikipedia.org/wiki/Cubic_equation

* http://mathworld.wolfram.com/CubicFormula.html