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Thread: Ti-83 difference of squares

  1. #1
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    Ti-83 difference of squares

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

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

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

    The first step I attempt to get $\displaystyle (x^4+\frac{1}{4})(x^4+\frac{1}{4}) $ by putting this into the calculator$\displaystyle (\frac{1}{16})^{(\frac{1}{4})}$ 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

    $\displaystyle 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 $\displaystyle 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 $\displaystyle \frac{1}{16}^{\frac{1}{2}}$ into your calculator and it will yield$\displaystyle \frac{1}{4}$
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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.

    $\displaystyle a^2-b^2 = (a-b)(a+b)$

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

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

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

    then applying the rule.

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