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### Factoring out a Fraction's Fifth Roots

```Date: 10/17/2011 at 07:12:31
From: Chris
Subject: 1/[(3*(8^(1/5))- 2*(4^(1/5)) -16^(1/5)- 2^(1/5)+3]

How do you factor out these fifth roots?

1/[(3*(8^(1/5)) - 2*(4^(1/5)) - 16^(1/5) - 2^(1/5) + 3]

I've already tried to change all the numbers inside the fifth root into
powers of two, i.e.,

8 = 2^3,
4 = 2^2,
16 = 2^4.

And I also let x = 2^(1/5).

But I still can't find an answer.

Thanks

```

```
Date: 10/20/2011 at 22:26:39
From: Doctor Vogler
Subject: Re: 1/[(3*(8^(1/5))- 2*(4^(1/5)) -16^(1/5)- 2^(1/5)+3]

Hi Chris,

Thanks for writing to Dr. Math.

Let's work our way up to a fifth root by starting with a simpler root, as
in this expression:

a + b*sqrt(q)

To rationalize that denominator, you multiply the numerator and the
denominator by the *conjugate* of your number. That's what you get by
replacing sqrt(q) with the other square root of q, namely -sqrt(q):

a - b*sqrt(q).

Working up from square roots, what if you have cube roots, instead? Then
there are two other conjugates of your number.

One replaces      cbrt(q)      by      w*cbrt(q)

The other replaces      cbrt(q)      by      w^2*cbrt(q)

Here, w^2 + w + 1 = 0 is the complex number

w = (-1 + sqrt(-3))/2.

This also has

w^3 = 1.

(You'll notice that those are the other two complex cube roots of q. These
conjugates are determined by the three complex embeddings of the number
field you mentioned, Q(cbrt(q)), into the complex numbers.)

In other words, the other two conjugates of this ...

a +   b*cbrt(q) + c*cbrt(q)^2

... are these two complex numbers:

a + b*w*cbrt(q) + c*w^2*cbrt(q)^2
a + b*w^2*cbrt(q) + c*w*cbrt(q)^2

The product of those two numbers is a real number (assuming that a, b, c,
and cbrt(q) are real numbers), so the w's go away.

(a + b*w*cbrt(q) + c*w^2*cbrt(q)^2)*(a + b*w^2*cbrt(q)
+ c*w*cbrt(q)^2)

= (a^2 - c*b*q) + (c^2*q - b*a)*cbrt(q)
+ (b^2 - c*a)*cbrt(q)^2.

And when you multiply the original expression a + b*cbrt(q) + c*cbrt(q)^2
by the above number, then you get a rational number, so you can
rationalize a denominator of a + b*cbrt(q) + c*cbrt(q)^2 by multiplying
the numerator and denominator by the above product.

You can do the same thing for fourth roots and fifth roots. In your case,
you have several powers of 2^(1/5). You need to multiply the numerator and
the denominator by the other four conjugates of your number. Each one is
gotten from your denominator by replacing each ...

2^(1/5)

... by

2^(1/5)*w^i

Here, i is an integer between 1 and 4, and w is a primitive fifth root of
1, so w satisfies the polynomial

w^4 + w^3 + w^2 + w + 1 = 0

This also implies that w^5 = 1.

So that means that when you're computing the conjugate that takes 2^(1/5)
to 2^(1/5)*w^2, this ...

8^(1/5) = (2^(1/5))^3

... turns into this:

(2^(1/5)*w^2)^3 = (2^(1/5))^3*w^6
= 8^(1/5)*w^6.

And so on.

Of course, you are going to have to multiply out a lot of terms, but
that's how it's done.

and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Square & Cube Roots

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