History Of Polynomial Equations
Cubic - Page One


 

To solve the general cubic, it is reasonable to begin by attempting to eliminate the a2 term by making a substitution of the form
z == x - lambda

Using
lambda = a2 / 3
will eliminate x2 and give

x^3 + 3 * (3a1-a2^2)/9 * x - 2 (9 a1 a2 - 27 a0 - 2a2^3)/54 = 0

[after some algebraic manipulation <vbg>]

1 - History
2 - Quadratics
3 - Cubic
4 - Quartic
5 - Quintic
6 - Appendix


 

Defining
p == (3a1-a2^2)/3
(9 a1 a2 - 27 a0 - 2a2^3)/27
then allows to be written in the standard form
x^3 + px = q

The simplest way to proceed is to make Vičta's substitution
x = w - p/(3w)

which reduces the cubic to the equation
w^3 - 9^3 / (27 w^3) - q = 0
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by thomas m. bösel @ www.vimagic.de for University Of Adelaide - History Of Mathematics 2002