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

  1. Dec 1, 2007 #1
    The book claims the method for the factorization is:

    [tex]x^n - c^n = (x^\frac{n}{2} - c^\frac{n}{2})(x^\frac{n}{2} + c^\frac{n}{2})[/tex]

    However, one example uses this:

    [tex]x^6 - \frac{1}{64} = x^6 - (\frac{1}{2})^6 = (x^3 - \frac{1}{2})(x^3 + \frac{1}{2})[/tex]

    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. jcsd
  3. Dec 1, 2007 #2

    rock.freak667

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    because it should be better written as

    [tex]x^6 - \frac{1}{16} = (x^3)^2 - (\frac{1}{4})^2 [/tex]
    to show how exactly it is a difference of two squares.
     
  4. Dec 1, 2007 #3

    James R

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    Note that

    [tex](\frac{1}{2})^6 = \frac{1}{64}[/tex]

    and not 1/16.
     
  5. Dec 1, 2007 #4
    Thanks. I copied it wrong.
     
  6. Dec 1, 2007 #5

    VietDao29

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

    [tex]\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}[/tex]
     
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