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

[Dave L. Renfro]

I've seen the identity

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)

applied in many ways (solving cubic equations,
obtaining the cubic arithmetic and geometric mean
inequality, showing that the collection of nonzero
rational linear combinations of {1, 2^(1/3), 2^(2/3)}
is closed under division, etc.), and it also makes frequent
appearances in advanced algebra texts from the early 1800s
to the early 1900s.

Does anyone know of some early appearances and uses
of this identity? By "early", I mean before the early
1800s. Also, has this identity, or something very
closely connected to it, been given a name at some
time in the literature (regardless of whether the name
was widely known at the time or whether the name survived
to the present time)?

Dave L. Renfro

[juandiego]

The identity would probably be known to Lagrange from
his extensive study of algebraic equations
If w is a primitive cubic root of unity then
a^3 + b^3 + c^3 - 3abc is the constant term
of the polynomial satisfied by a +bw +cww

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

(a+b+c)(a +bw +cww)(a +bww +cw).
This and other similar identities occur when
symmetrical functions of the roots of polynomial
equations are calculated
I know Newton studied symmetric functions of roots.
He may have been aware of this identity.

(a +bw +cww) is a special form of Eisenstein cubic
integrers and a^3 + b^3 + c^3 - 3abc ist its norm.
a^3 + b^3 + c^3 - 3abc can also be referrred to a
a termary cubic form.

[Bill Dubuque]

It's a special case of the the well-known identities expressing
power sums in terms of elementary symmetric polynomials, i.e. it's

p3 - 3 e3 = e1 p2 - e2 p1

Gauss gave an algorithm to rewrite any symmetric poly as a poly
in the elementary symmetric poly's. Nowadays this may be viewed
simply as a very special case of elimination via Grobner bases.

This easily generalizes to the ring of invariants of a
finite matrix group G < GL(n,k), e.g. see chapter 7 of
Cox, Little, O'Shea: Ideals, Varieties and Algorithms.

They claim (p. 314) that in Gauss' proof is the earliest
known explicit statement of the lexicographic order.

--Bill Dubuque

[Dave L. Renfro]

Bill Dubuque wrote (in part):

> It's a special case of the the well-known identities expressing
> power sums in terms of elementary symmetric polynomials,
> i.e. it's
>
> p3 - 3 e3 = e1 p2 - e2 p1

Thanks for the terms (mentioned later in your comments).
As for the above, I haven't looked through an extensive
collection of papers I have on cubic equations yet (probably
over 1000 that are essentially unsorted and in several boxes),
but I've seen the above in Arthur Engel's "Problem-Solving
Strategies", 1998 [Chapter 10, Problem 1, pp. 254 & 258],
and figured there was more to the story than what he gave.

I'm thinking of writing a short expository paper on the
applications of this identity, along with ways of obtaining
the factorization (I know several, one of which involves
expanding a certain 3-by-3 determinant) and its various
appearances in old university entrance exams.

Here are some more examples besides what I listed:

6 is a factor of k^3 + m^3 + n^3 iff 6 is a factor of k + m + n

A triangle with vertices a, b, c in the complex plane is
equilateral if and only if a^2 + b^2 + c^2 - ab - bc - ca = 0.

The fact that a + b + c = 0 implies a^3 + b^3 + c^3 - 3abc = 0
can be used to obtain an equation for the cubes of the roots of
px^3 + qx^2 + rx + s = 0

I'd be very interested in other uses of this identity that anyone
here might know about.

Problem on M'Cay's Junior Freshmen Mathematics Dublin Examination
Paper for 1885: Given x = 2a - b - c, y - 2b - c - a, and
z = 2c - a - b, find the value of x^3 + y^3 + z^3 - 3xyz.

Problem on Bernard's Arithmetic and Algebra Entrance Examination
for the Trinity Term for 1885: If a = 5, b = 4, c = 3, find the
numerical value of sqrt[(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)].

Problem on W. R. Roberts' Additional Examination for High Places
in Geometry and Algebra for the Trinity Term for 1885: If a+b+c = 0,
show that a^3 + b^3 + c^3 - 3abc = 0.

Dave L. Renfro

[Dave L. Renfro]

juandiego wrote (in part):

> (a +bw +cww) is a special form of Eisenstein cubic
> integrers and a^3 + b^3 + c^3 - 3abc  ist its norm.
> a^3 + b^3 + c^3 - 3abc can also be referrred to a
> a termary cubic form.

Thanks, I'll look-into/research this term some. See
my reply to Bill Dubuque for some more general comments.

Dave L. Renfro

[juandiego]

On 24 Apr, 15:12, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
> Bill Dubuque wrote (in part):

> I'm thinking of writing a short expository paper on the

> applications of this identity, along with ways of obtaining
> the factorization (I know several, one of which involves
> expanding a certain 3-by-3 determinant)

Yes, a^3 + b^3 + c^3 - 3abc is the circulant or determimant
of the circulant matrix
{a,b,c; c,a,b; b,c,a}

You may find some historical references in Chrystal's
Algebra.

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)
The above form is a special case of the determinant of
{a,kb,kc; c,a,kb;b,c,a}

You will find similar identities in Ed Barbeau's
"Pell's Equation"
a^3 + kb^3 + kkc^3 - 3kabc = 1
is a generalization of the standard
Pell equation x^2-ky^2 =1

[Dave L. Renfro]

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?

Dave L. Renfro

[juandiego]

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.