The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Rationalizing a Denominator with Multiple Cube Roots

Date: 04/22/2011 at 05:08:13
From: Steve (but you may call me stevie)
Subject: Rationalize a denominator with multiple cube roots

Hi Dr. Math,

Or may I call you 'doc'?

I have the following problem:

Given this fraction of rational numbers a, b, c, and q in Q:

   1/(a + b*CBRT(q) + c*CBRT(q)^2)

I was wondering if it is possible to express this general formula without
non-rational numbers in the bottom of the fraction -- in other words, get
rid of the cube roots in the denominator.

I know this should be possible, as we know from basic algebra that this is
isomorphic with Q(CBRT(q)):

   Q[x]/(x^3 - q) 

Moreover, all elements of the first one are of the form 

   q1 + q2*X + q3*X^2

With this knowledge, it should be clear that what I am trying to achieve
should be possible. Alas, I still fail to get the right form of that
general case.

I've tried to use a formula involving (a + b)(a^2 + ab + b^2), but this
does not seem to work on this case.

Could you help me out, doc?

Date: 04/22/2011 at 19:42:19
From: Doctor Vogler
Subject: Re: Rationalize a denominator with multiple cube roots

Hi Stevie,

Thanks for writing to Dr. Math. 

Before we take up your question, let's start with a simpler root. 

To rationalize a denominator with a square root, like ...

  a + b*SQRT(q),

... you multiply the numerator and the denominator by the *conjugate* of
your number. The conjugate is what you get by replacing SQRT(q) by the
other square root of q, namely -SQRT(q):

  a - b*SQRT(q).

Well, in the case of cube roots, there are two other conjugates of your
number. One replaces CBRT(q) by w*CBRT(q). The other replaces CBRT(q) by
w^2*CBRT(q), where ...

   w^2 + w + 1 = 0 

... is the complex number 

   w = (-1 + SQRT(-3))/2,

... which also has w^3 = 1. (Notice that those are the other two complex
cube roots of q. These conjugates are determined by the three complex
embeddings of the number field you mentioned, Q(CBRT(q)), into the complex

In other words, the other two conjugates of ...

  a + b*CBRT(q) + c*CBRT(q)^2

... are the two complex numbers ...

  a + b*w*CBRT(q) + c*w^2*CBRT(q)^2

... and ...

  a + b*w^2*CBRT(q) + c*w*CBRT(q)^2.

Assuming that a, b, c, and CBRT(q) are real numbers, as you have, the
product of those two numbers is also a real number, so the w's go away:

    (a + b*w*CBRT(q) + c*w^2*CBRT(q)^2)
  * (a + b*w^2*CBRT(q) + c*w*CBRT(q)^2)
  = (a^2 - c*b*q) + (c^2*q - b*a)*CBRT(q) 
                  + (b^2 - c*a)*CBRT(q)^2

Well, if you multiply the numerator and the denominator of your fraction
by this number, then the denominator will be a polynomial in a, b, c, and
q, namely:

  a^3 + (b^3 - 3*c*b*a)*q + c^3*q^2.

That is also the norm of your denominator as an element of Q(CBRT(q)).

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further

- Doctor Vogler, The Math Forum 
Associated Topics:
High School Number Theory
High School Square & Cube Roots

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.