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```
Date: 11/28/2001 at 17:26:12
From: Anna

I understand how you use the square root, but how do you use the cubed
root?

The problems are:

sqrt(24)
---------------------------------------
sqrt(2) + sqrt(3) + sqrt(5)

6
--------------------------
1 - cubedroot(3)

12
------------------------------------
cubedroot(4) - cubedroot(2) + 1

Thanks a million!
```

```
Date: 11/29/2001 at 10:29:04
From: Doctor Peterson

Hi, Anna.

The basic idea used in this kind of problem is to use a factorization
like these:

a^2 - b^2 = (a - b)(a + b)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

Check out those last two if you aren't familiar with them.

When a and b are square or cube roots, these formulas tell us we can
multiply a-b or a+b by a certain polynomial so that the product will
involve only the square or cube of a and b, eliminating the radical.
You can do the same with even higher powers; just divide a^n - b^n by
a - b, for example, to find the other factor.

Your first example involves only square roots, but is tricky because
there are three of them:

sqrt(24)
---------------------------
sqrt(2) + sqrt(3) + sqrt(5)

All we can do is try something; so let's take a = sqrt(2) + sqrt(3)
and b = sqrt(5). If you multiply numerator and denominator by a - b,
which is [sqrt(2) + sqrt(3)] - sqrt(5), you will find that the
denominator becomes very simple, and it doesn't take much more work to
finish rationalizing the denominator. Other choices may work just as
well; try them to familiarize yourself with how these things work.

The next one uses the cube root:

6
-----------
1 - cbrt(3)

(Note that it is "cube root", not "cubed root," which would suggest
you are cubing the square root. I shortened your abbreviation to my
own favorite.)

Here you can use my second formula with a = 1 and b = cbrt(3),
by multiplying numerator and denominator by
a^2 + ab + b^2 = 1 + cbrt(3) + cbrt(9). See what happens!

Finally,

12
---------------------
cbrt(4) - cbrt(2) + 1

Here we have both complications; it involves cube roots, and there are
three terms in the denominator. My first thought was to do the same
thing I did for the first problem, taking part of the denominator as a
and part as b, and seeing what happened; you should try that for the
experience. You may get a surprise! But working with that made me
realize that cbrt(4) is the square of cbrt(2), so that if we just let
a = cbrt(2), the denominator is a^2 - a + 1. Look familiar? See if you
can use that as a clue.

These are a little challenging, but once you get past these, you'll be
able to handle most of what they can give you.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/04/2001 at 18:28:33
From: Anna

Thank you. I think I get it.
```
Associated Topics:
High School Exponents

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