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Math Help - Ti-83 difference of squares

  1. #1
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    Ti-83 difference of squares

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

    I like to use the calculator to double check myself, but for some reason I cannot produce the correct answer for the problem below.

    x^8-\frac{1}{16}=(x^4+\frac{1}{4})(x^4+\frac{1}{4})
    =(x^2-\frac{1}{2})(x^2+\frac{1}{2})(x^4+\frac{1}{4})

    The first step I attempt to get (x^4+\frac{1}{4})(x^4+\frac{1}{4}) by putting this into the calculator  (\frac{1}{16})^{(\frac{1}{4})} which results in .5 or 1/2. How do I keep on missing the first step to the solution which is the 1/4?
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  2. #2
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    Quote Originally Posted by allyourbass2212 View Post

    x^8-\frac{1}{16}=(x^4+\frac{1}{4})(x^4+\frac{1}{4})
    I don't think this is correct. I believe it is x^8-\frac{1}{16}=(x^4-\frac{1}{4})(x^4+\frac{1}{4})
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  3. #3
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    Sorry that was a typo, and my question is where does the come from, and from what operation exactly in the calculator?
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  4. #4
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    Hello Allyourbass,



    What squared is 16? Simply input \frac{1}{16}^{\frac{1}{2}} into your calculator and it will yield  \frac{1}{4}
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  5. #5
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    Quote Originally Posted by allyourbass2212 View Post
    Sorry that was a typo, and my question is where does the come from, and from what operation exactly in the calculator?
    this comes from the difference of 2 squares rule.

     a^2-b^2 = (a-b)(a+b)

    x^8-\frac{1}{16}

    =(x^4)^2-\frac{1}{4^{-2}}

     =(x^4)^2-(\frac{1}{4^{-1}})^2

    then applying the rule.

     =(x^4-\frac{1}{4})(x^4+\frac{1}{4})
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