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Cardan's Cubic Method

Date: 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
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Associated Topics:
College Calculus
High School Calculus

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