Cardan's Cubic MethodDate: 7/19/96 at 23:36:50 From: Peter W. Messer Subject: Cardan's Cubic Method; an Algebraic Mess Dear Dr. Math, Cardan's well known formula solution of cubic equation x^3 - 3x^2 - x - Sqrt[2] = 0 gives us a single "monstrous" algebraic real root which appears to defy further simplification to the expected simple root x = 2 + Sqrt[2]. For the class of cubic polynomial equations that contain Sqrt[n] in their coefficients are there general algebraic methods for finding the maximally simplified roots without resorting to guessing or "special tricks" that only apply to some cases? Any help on this difficult (?) problem will be appreciated. Peter Messer Date: 7/20/96 at 9:56:1 From: Doctor Anthony Subject: Re: Cardan's Cubic Method; an Algebraic Mess If we put y=x-1 or x = y+1 the equation becomes: (y+1)^3 -3(y+1)^2 - (y+1) - sqrt(2) = 0 y^3 + 3y^2 + 3y + 1 - 3y^2 - 6y - 3 - y - 1 - sqrt(2) = 0 y^3 - 4y - 3 - sqrt(2) = 0 Compare with the standard equation y^3 + 3Hy + G = 0 we see that H = -4/3 and G = -3 - sqrt(2) With Cardan's solution, if p, q are chosen so that p^3+q^3 = G and pq = -H then the roots of the cubic will be: -p-q, -wp-(w^2)q, -(w^2)p-wq where w = complex cube root of 1. But p^3, q^3 will be the roots of the quadratic t^2 - Gt - H^3 = 0, Having found p^3 and q^3 you then require -(p+q) for the real root, and the algebra involved in expressing this in radicals would be, as you have observed, horrendous. An alternative is to remember that p^3 + q^3 does factorise to (p+q)(p^2-pq+q^2). We have p^3+q^3 = G = -3 - sqrt(2) and this factorises to: -(2+sqr(2)){2-(1/2)sqrt(2)}, and we can check that with pq = 4/3 we can identify the bracket (2+sqrt(2)) with -(p+q) and p^2-pq+q^2 with the second bracket. This shows that the real root is 2+sqrt(2), but I should certainly not recommend it as a general method. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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