Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Simplifying with Cube Roots


Date: 01/22/97 at 14:45:39
From: Grossman
Subject: Cube root

Hi there,

I have an incredibly hard (I think) algebra problem.  I would really 
appreciate it if you could show me how to get the answer to this 
problem:

   cube root(3) / [cube root(6) + cube root(3)]  

I need the answer simplified, and you can't just use a calculator (I 
tried).  

Thanks in advance,
Adam


Date: 01/22/97 at 15:27:59
From: Doctor Wilkinson
Subject: Re: Cube root

You're right.  This is a pretty hard problem.

First of all, you can simplify it a bit by dividing the numerator and
denominator by the cube root of 3.  This makes it:

1/(1 + cube root(2))

Now to get the cube root out of the denominator, we're going to use 
the identity:

1 + x^3 = (1 + x) (1 - x + x^2)

which you may remember from factoring.  If you apply this with 
x = cube root(2), you get:

1 + cube root(2)^3 = (1 + cube root(2))(1 - cube root(2) + cube root(2)^2)

or

1 + 2 = (1 + cube root(2)) (1 - cube root(2) + cube root(4))

or

1 + cube root(2) = 3/(1 - cube root(2) + cube root(4))

or

1/(1 + cube root(2)) = (1 - cube root(2) + cube root(4))/3

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 01/22/97 at 16:05:49
From: Doctor Math
Subject: Re: Cube root

Hi -

Here's a hint about what you can do with this:

The cube root of 3 is a number.  In fact, it's the positive number 
that, when you raise it to the third power, gives you three.  So the 
cube root of 8 is 2, because 2 cubed (also written as 2^3 or 2*2*2) is 
8.  They're a lot like square roots, and if you know how to work with 
those then you're well off for working with cube roots.  The cube root 
of three can be written as 3^(1/3).  That's three to the one-third 
power.  

If you've got anything that looks like (ab)^c, that equals a^c * b^c.  
So in this case, in the denominator we've got the cube root of 6, 
which we can write as 6^(1/3).  By the rule I mentioned above, that's 
equal to 2^(1/3) * 3^(1/3), which is the cube root of 2 times the cube 
root of 3.  

Using all of the stuff I just talked about, we can change this:

          cube root(3) / [cube root(6) + cube root(3)]  

to this:
                           3^(1/3)
                     -------------------
                      6^(1/3) + 3^(1/3)

The REASON we can use that one-third power to represent a cube root 
is this: let's say you want the cube root of x.  If you take the number 
x^(1/3) and cube it, you get:

          (x^(1/3))^3 = x^((1/3) * 3) = x^1 = x

So that means x^1/3 must be the cube root of x.

Anyway, once you have it in this form, you can write 6^(1/3) as 
(2*3)^(1/3), and that's the same as 2^(1/3) * 3^(1/3).  So we have this:

                             3^(1/3)
                  -----------------------------
                   2^(1/3) * 3^(1/3) + 3^(1/3)

And then you can simplify by factoring out a 3^(1/3) from the bottom:

                          3^(1/3)
                  -----------------------
                   3^(1/3) (2^(1/3) + 1)

I bet you see what to do next!

-Doctor Ken,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Exponents
Middle School Algebra
Middle School Exponents

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/