# Can I get a little help? (Factoring difference of 2 sqaures).

1. Dec 1, 2007

### Holocene

The book claims the method for the factorization is:

$$x^n - c^n = (x^\frac{n}{2} - c^\frac{n}{2})(x^\frac{n}{2} + c^\frac{n}{2})$$

However, one example uses this:

$$x^6 - \frac{1}{64} = x^6 - (\frac{1}{2})^6 = (x^3 - \frac{1}{2})(x^3 + \frac{1}{2})$$

My question is, why isn't 1/2 raised the 3rd power like the method tells you to do?

Any information would be greatly appreciated!

Last edited: Dec 1, 2007
2. Dec 1, 2007

### rock.freak667

because it should be better written as

$$x^6 - \frac{1}{16} = (x^3)^2 - (\frac{1}{4})^2$$
to show how exactly it is a difference of two squares.

3. Dec 1, 2007

### James R

Note that

$$(\frac{1}{2})^6 = \frac{1}{64}$$

and not 1/16.

4. Dec 1, 2007

### Holocene

Thanks. I copied it wrong.

5. Dec 1, 2007

### VietDao29

Well, because the book misprinted it. :) You can easily expand all the terms on the RHS out to get:

$$\left( x ^ 3 - \frac{1}{2} \right) \times \left( x ^ 3 + \frac{1}{2} \right) = x ^ 6 - \frac{1}{4} \neq x ^ 6 - \frac{1}{64}$$