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Rationalizing a Cubic Root Denominator

Date: 12/06/97 at 16:16:46
From: Eric Walker
Subject: Rationalizing a cubic root denominator

I am trying to rationalize the equation:


so far I have tried to multiply it by  cubicrt(9)+2

but that obviously didn't work. I also tried using my graphing 
calculator to find a general way of solving for
but I couldn't figure it out. Please help!

Eric Walker

Date: 12/16/97 at 10:38:21
From: Doctor Mark
Subject: Re: Rationalizing a cubic root denominator

Hi Eric,

Sometimes you need to find the correct analogy to help you see what to 
do. You tried to rationalize the denominator by proceeding in analogy 
with rationalizing the denominator when you have a square root in the
denominator. You remember how this goes:

To rationalize the denominator of

   B - sqrt(C)

you multiply top and bottom of the fraction by B + sqrt(C). Why does 
that work?

It works because:

   (m + n)(m - n) = m^2 - n^2

So if n = sqrt(C), n^2 will just equal C, and the square root will 
have disappeared (on the bottom of the original fraction).

Said another way, to "rationalize" B - sqrt(C), we need to multiply 
it by some quantity that effectively squares the "sqrt(C)," giving 
us a C. The quantity which does that is B + sqrt(C).

So now our question is this: if we have, instead, B - cubrt(C), what 
would we have to multiply this by to effectively cube the "cubrt(C)," 
giving us back a C?

We can get a hint at how that goes by looking at the product involving 
m and n above, but turning it around backwards, as a factoring 

   m^2 - n^2 = (m - n)(m + n).

Then if m or n (or both!) are square roots, we can "get rid of the 
square roots" by multiplying (m + n) by (m - n), or by multiplying 
(m - n) by (m + n), depending on which one we have originally.

If we want to get rid of *cube* roots, we might want to look at the
analogous quantity, but using cubes, instead of squares, of m and n.   
So we might try looking at a difference of cubes:

   m^3 - n^3 = ?

or a sum of cubes:

   m^3 + n^3 = ?

If m or n (or both) are cube roots, then m^3 - n^3 and m^3 + n^3 will 
*not* involve cube roots!

Do you know how to factor these?  You can check [see the end for a way 
of figuring this out] that these are the factorizations:

   m^3 - n^3 = (m - n)(m^2 + mn + n^2)

   m^3 + n^3 = (m + n)(m^2 - mn + n^2)

So we see that if we have a quantity which is a sum of two terms 
(m + n), where either m or n, or both, is a cube root, multiplying by 
(m^2 - mn + n^2) will give us m^3 + n^3, which will *not* involve cube 
roots, and similarly for a quantity that is the difference of two cube 

For the problem you have,

   cubrt(9) - 2

m is like cubrt(9), and n is like 2, so multiply the top and bottom of 
this fraction by

   m^2 + mn + n^2 = (cubrt(9))^2 + (cubrt(9)(2) + (2)^2

                  = (cubrt(9^2)) + 2cubrt(9) + 4 ==  Q.

Then you will get a fraction with Q on the top, and

(m - n)(m^2 + mn + n^2) = m^3 - n^3 = (cubrt(9)^3 - (2)^3 = 9 - 8 = 1

on the bottom; i.e., your answer will be

   -------------  =  (cubrt(9^2)) + 2cubrt(9) + 4
   cubrt(9) - 2

Similar considerations would apply if you had cubrt(9) + 2 on the 

Overall, a good question, Eric.

And now I have one for you: how would you rationalize the denominator 
if the denominator looked like

   A^1/4 - B^1/4  ?

   (Hint: how do you factor m^4 - n^4?)

As promised, here is how you might have guessed the factorization for 
a difference or a sum of cubes.

The Remainder theorem can be interpreted as saying that if a is a 
solution to f(x) = 0, then x - a is a factor of f(x). If we apply
this to m^3 - n^3, then (with a little wiggling around), we see 
that if m = n, then m^3 - n^3 = 0, and hence (m - n) is a factor of 
m^3 - n^3. We can then find the factorization of m^3 - n^3 by just 
dividing m^3 - n^3 by m - n.

A similar argument holds for m^3 + n^3:  this equals 0 if m = - n, and 
so (m - (-n)) = (m + n) is a factor of m^3 + n^3. We can then find the
factorization just as before.

Hope this all makes sense to you Eric; if any part of it is unclear, 
write back to me and I'll have another go at it.

-Doctor Mark,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Exponents

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