Factoring difference of sixth powers: why not cube the base?

[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!

 

[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.

 

[James R]

Note that

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

and not 1/16.

 

[Holocene]

Note that

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

and not 1/16.

Thanks. I copied it wrong.

 

[VietDao29]

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!

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

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