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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:

1
------------
cubicrt(9)-2

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

but that obviously didn't work. I also tried using my graphing
calculator to find a general way of solving for
1
------------
cubicrt(a)-b
Thanks,

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

A
-----------
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
problem:

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
roots.

For the problem you have,

1
-------------
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

1
-------------  =  (cubrt(9^2)) + 2cubrt(9) + 4
cubrt(9) - 2

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

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!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents

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