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Cubic_Equation

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					                               Mathematics Department
                               Solving a Cubic Equation

The quadratic formula for the general degree two equation is one of the most familiar
equations in mathematics. A formula exists for solving the general cubic equation but
it is not easily quotable. I was surprised at discovering a non-standard derivation of
the formula in an article in the library whilst I was looking for something else. It goes
as follows:

The solutions of x 3  px  q  0 are x  (b  c) 2 x  (b  c), x  (b  c)

                        1 i 3                   q  q 2  (4 p 3 27) 
where  3  1 and             and b  3                               and c   p 3b .
                          2                                2           
                                                                       
Proof
We wish to solve the cubic ax 3  bx 2  cx  d  0 . Using a linear change of variable,
this cubic can be reduced to:
                              x 3  px  q  0 .      (1)

Now the following identity holds (a, b and c are arbitrary complex constants):

            (a  b  c)(a  b  c)(a  b  c)   (a 3  3abc  b 3  c 3 )

Replacing a with x, this produces:

        (x  b  c)( x  b  c)( x  b  c)   ( x 3  3xbc  b 3  c 3 )    (2)

Now we can recognise that the right side of (2) is essentially the same as what appears
in (1), provided the following hold:

                                   3bc  p            (3)
                                  b c  q
                                    3    3
                                                         (4)

So given the values of p and q, we need only determine a b and c satisfying these
relations. (2) will then give three linear factors and hence three solutions. We can
rewrite the equations (3) and (4) as:
                                                  p3
                                        b 3c 3 
                                                  27
                                     b c  q
                                      3      3


                                                                             3 
One can see that b 3 and c 3 are roots of the quadratic equation x 2  qx   p    0.
                                                                             27 
                                             1

                 q  q 2  (4 p 3 27)  3                                  
                                         . Let b  3  q  q  (4 p 27) 
                                                                2       3

These roots are 
                          2                                     2         
                                                                          
and by (3) c   p 3b .
                                     (b  c)
Hence the solutions of (1) are x             , x  (b  c), x  (b  c)
                                         

JL/WGN
24 May, 2011

				
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