The Cubic Equation

An article from a very informed friend R.W.D Nickalls on his method of solving Cubics

There exists an exact solution to this equation and this solution will be shown in due course.
There existed a time when the solution to the Cubic was not known and for hundreds of years people have
searched for the solution. Then some Italians from their University solved the Cubic.

Proof follows
 
We start by reducing the equation to the form v3 + v + h = 0



Reducing the Cubic
Given Ax3 + Bx2 + Cx + D = 0    eq1

Let x = z + m , so x = f(z)

Then x2 = z2 + 2zm + m2

And x3 = z3 + 3mz2 + 3zm2 + m3

Divide eq1 by A and we get x3 + ax2 + bx + c = 0        - eq2


Now we substitute into eq2 and we get


(z3 + 3mz2 + 3zm2 + m3 ) + a(z2 + 2zm + m2) + b(z+m) + c = 0


which is z3 + ( 3m + a )z2 + ( 3m2 + 2am + b )z + (m3 + am2 + bm + c) = 0


Now we get rid of the z2 term by letting 3m+a = 0 so that m = -a/3


So now we are left with z3 + ( a2/3 - 2a2/3 +b)z + ( -a3/27 + a3/9 -ba/3 +c) = 0


Which reduces to z3 + ( b - a2/3)z + ((2/27)a3 - ab/3 + c) = 0


Let b - a2/3 = f2 and let (2/27)a3 - ab/3 + c = g


Then we have z3 + f2z + g = 0


Now let z = fv so that z = f(v) where z3 = (fv)3

Therefore f3v3 + f3v + g = 0 


Divide by f3 and we have v3 + v + g/f3 = 0


Now let h = g/f3 and we have v3 + v + h = 0 and we are finished with reducing the Cubic






The roots of the reduced Cubic

Let h = -K

We now have reduced our original equation to v3 + v  - K =0  -eq#

Solving this equation will prove that an exact solution to the Cubic exists



Now let v = p1/3 + q1/3 so v = f(p,q)

Therefore v3 = p + 3p2/3q1/3 + 3p1/3q2/3 + q

We substitute into eq# and we have (p+q) +3(pq)1/3 [ p1/3 + q1/3] + (p1/3 + q1/3) -K = 0

Now we let p + q = K and we are left with (p1/3 + q1/3)(1 + 3[pq]1/3) = 0

So 1 + 3[pq]1/3 = 0 as p1/3 + q1/3 = v <> 0 if K<>0

Therefore pq = -1/27 , so p = - 1/[27q]

We know that p+q = K , so p = K - q = -1/[27q]

So 27Kq - 27q2  +1 = 0 which is q2 - Kq - 1/27 = 0

We see that we have reduced our equation to that of a Quadratic which can be easily solved

So q = [K +/- sqrt(K2 + 4/27)]/2 = K/2 +/- sqrt(K2/4 + 1/27) 

We use root q = K/2 + sqrt(K24 + 1/27)

Therefore p = K - q = K/2 - sqrt(K2/4 + 1/27)

So v = p1/3 + q1/3  = [ K/2 - sqrt(K2/4 + 1/27) ]1/3 +  [ K/2 + sqrt(K2/4 + 1/27) ]1/3
 
We see therefore that we have a solution for the equation in v 
and thus also for x as z = fv and x = z + m

We therefore have proved an exact solution to the Cubic



Example using the reduced Equation

We demonstrate by using x3 + x - 10 = 0 which by inspection show that x = 2 is a root.

So K =10

Now from our formula we have x = [5 - sqrt(25 + 1/27)]1/3 + [5 + sqrt(25 + 1/27)]1/3

So x = (5-5.003702333..)1/3 + (5+5.003702333..)1/3 = -0.154700539.....+ 2.154700539... = 2


So our formula works and all is well 





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