Re: Factorization of a^3 + b^3 + c^3 - 3abc

On 24 Apr, 18:10, "Dave L. Renfro" <renfr...@xxxxxxxxx> wrote:
juandiego wrote (in part):

In your original post you say that a^3 + b^3 + c^3 - 3abc
can be used to show that "the collection of nonzero

rational linear combinations of {1, 2^(1/3), 2^(2/3)}
is closed under division, etc.)".

In this case do you mean  a^3 + 2b^3 + 4c^3 - 6abc ?
This is the norm of numbers of the form
a + b2^(1/3)+ c2^(2/3)

Let's see ...

We want a rationalizing factor for r + s*2^(1/3) + t*2^(2/3).
If I let a = r, b = s*2^(1/3), and c = t*2^(2/3), then
multiplying r + s*2^(1/3) + t*2^(2/3) by the real-quadratic
factor of a^3 + b^3 + c^3 - 3abc gives

a^3 + b^3 + c^3 - 3abc

= r^3 + 2s^3 + 4t^3 - 3[r][s*2^(1/3)][t*2^(2/3)]

= r^3 + 2s^3 + 4t^3 - 6rst,

which is rational when each of r, s, t is rational.

This seems to work, unless I'm missing something. Perhaps
the "norm of the numbers of the form ..." has additional
properties beyond that of providing a rationalizing factor?

Yes, I thought you meant that
a^3 + b^3 + c^3 - 3abc was the "rationalizer",
when obviously it is a^3 + 2b^3 + 4c^3 - 6abc.
But of course all you need to do is nake a substitution.

Your approach must then also work for {1,r,rr}
where r^3 = k, k rational, but not a rational cube.
.