Complex Radicals

Date: 11/28/2001 at 17:26:12
From: Anna
Subject: Complex radicals

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
Subject: Re: Complex radicals

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
Subject: Re: Complex radicals

Thank you. I think I get it.