Slog about cardan

cardan by .
The derivation is long and messy. (You don't simply complete the cube.)

If you start with x^3+bx^2+cx+d=0, you first make the substitution
x = y -b/3. The result is a "reduced cubic equation" which has the form:
y^3+ py + q = 0. You then can find the roots of this by making another substitution y =z - p/(3z) which yields a 6 th degree equation in z:
z^6 +qz^3 -p^3/27 = 0. You solve this for z^3 by the quadratic equation ( since z^6 = (z^3)^2 ). The argument then gets more complicated and involves knowing about the imaginary cube roots of unity, etc., etc..
As I said, long and messy..

You can find it worked out in older math books on the theory of equations. I'm sure you can google it and find a Web site that has it worked out as well.

June 30, 2009 | 3:05 pm | Report abuse | Reply