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### Rationalizing Denominators with Cube Roots

```Date: 06/26/2008 at 16:38:42
From: Lea
Subject: Rationalizing Denominators

Dear Dr. Math,

How do you rationalize this denominator:

1
------------------------------
cubrt(6) + cubrt (9) + cubrt(4)

I find it hard to use sums of cubes.  I grouped cubrt(6) and cubrt(9)
so can use sums of cubes but the problem became more complicated.  Is
there a technique to rationalize this expression?

```

```
Date: 06/28/2008 at 11:41:34
From: Doctor Ali
Subject: Re: Rationalizing Denominators

Hi Lea!

Thanks for writing to Dr. Math.

So we have,

1
---------------------------
__       __       __
3 /      3 /      3 /
\/ 6  +  \/ 9  +  \/ 4

We have cube roots in the denominator, so we have to multiply
something by top and bottom to make cubes to cancel out the cube
roots.

Do you know this formula?

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

We can let

a = cbrt(6)
b = cbrt(9)
c = cbrt(4)

So, we have (a + b + c) in the denominator.  We can multiply top and
bottom by (a^2 + b^2 + c^2 - ab - ac - bc) to get rid of the cube
roots.  In other words, we should multiply top and bottom by,
__        __        __        __        __        __
3 /       3 /       3 /       3 /       3 /       3 /
S =  \/ 36  +  \/ 81  +  \/ 16  -  \/ 54  -  \/ 24  -  \/ 36

Does that make sense?

The numerator will be S itself.  The denominator will be

__        __        __        __________
3 / 3     3 / 3     3 / 3     3 /
\/ 6  +   \/ 9  +   \/ 4   -  \/ 6 x 9 x 4

Which is,

6  +  9  +  4  +  6  =  25

a rational number!

Does that make sense?

You might also want to take a look at:

Rationalizing Denominators
http://www.mathforum.org/library/drmath/view/53052.html

This is another example of rationalizing denominators with odd roots
in the denominator.

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials

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