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Expert: - 8/1/2010


Question
Good afternoon Scott,
I'm having some problems reducing the cubic and wondered if you could help at all.

    x3  + b x2  + c x + d = 0

The quadratic term disappears
Now we want to reduce the last equation by the substitution

       x = y + r

The cubic equation becomes:

  (1)   (y + r)3  + b (y + r)2  + c (y + r) + d = 0
<=>
  (2)    y3  + (3 r + b) y2  + (3 r2  + 2 r b + c) y + r3  + r2  b + r c + d = 0


My problem is- I cannot see how to get from (1) to (2).


The question I have to answer this for is:

We first make a composition of f with a linear function g(x) = x - c . For f(x) = x^3 + 6x^2 + 18x choose c so
that the composition f of g does not contain any term in x^2 . This is called the “reduced cubic” and is easier to deal
with than the original form. [3]

( do not know what 'choose c' means )

We shall now assume that the general cubic polynomial has been “reduced” so that it may be written as
f(x) = x^3 - 3ax + b,


Many thanks!
Anya

Answer
We take (y + r)^3  + b(y + r)^2  + c(y + r) + d = 0.
Now (y+r)^3 = y^3 + 3y^2r + 3yr^2 + r^3, b(y+r)^2 = by^2 + 2byr +2br^2, and c(y+r) = cy + cr.
The 1st term of the cube is y^3.
The 2nd term of the cube and the 1st term of the sqaure are 3y^2r + by^2, which is (3r+b)y^2.
The 3rd term in the cube, the 2nd term in the square, the the 1st term in the line make up
3yr^2 + 2byr + cy, which factors to (3r^2 + 2br + c)y.
The constants (with no y) can be grouped together in the same way.

I'm not sure how at this point, but if the right c is chosen, the square term might disappear.
I believe the correct way to choose it is to make 3r+b=0, and then get the value of c from there.  

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