The next stage up from the quadratic equation, the general cubic
equation has the form $ax^3+bx^2+cx+d=0.$
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:
Substitute $x=t-a/3$ giving
- $(t-a/3)^3+a(t-a/3)^2+b(t-a/3)+c=0$
That has eliminated the quadratic term, so we have
for suitable p and q.
Now substitute $t=u-\frac{p}{3u}$ and multiply by $u^3$ to get
- $u^6+qu^3+\frac{p^3}{27}=0$
Now we have a quadratic in $u^3$ so using the quadratic equation
formula we get two possible answers for $u^3.$ 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.
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