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### Simplifying Cube Roots in the Denominator

```
Date: 07/12/2001 at 08:39:18
From: pauline viray
Subject: Simplifying cube roots in denominator

I cannot simplify this:

The cube root of 3 all over the (cube root of 2) minus the (cube root
of 4).

Thanks!
```

```
Date: 07/12/2001 at 12:38:10
From: Doctor Jaffee
Subject: Re: Simplifying cube roots in denominator

Hi Pauline,

I assume that what you did was multiply the numerator and denominator
of your expression by cube root of 2 + cube root of 4. This process
would work fine if you were trying to rationalize sqrt(2) - sqrt(4),
but it doesn't work with cube roots.

The reason your method works with square roots is that any expression
in the form a^2 - b^2 can be factored to (a - b)(a + b), so if you
have an expression in the form a - b and you multiply it by a + b, you
end up with both terms being squared. In the problem above, however,
you want both terms cubed.

So, let's use the fact that a^3 - b^3 = (a - b)(a^2 + ab + b^2).

In your problem a = cube root of 2 and b = cube root of 4. Multiply
the numerator and denominator by a^2 + ab + b^2 and the resulting
denominator will be cube root of 2^3 - cube root of 4^3 = -2.

Give it a try and if you want to check your solution with me, write
back. If you are having difficulties, let me know and show me what you
have done so far, and I'll try to help you some more.

Good luck.
- Doctor Jaffee, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents

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