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Thread: Simplification of Fractional Expressions

  1. #1
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    Simplification of Fractional Expressions

    How do I simplify these fractional expressions?

    $\displaystyle (x + h)^-3 - x^-3 \over h$

    $\displaystyle \sqrt{1+(x^3-\frac{1}{4x^3})^2} $

    thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nathan02079 View Post
    $\displaystyle (x + h)^-3 - x^-3 \over h$
    $\displaystyle \frac{ (x + h)^{-3} - x^{-3}}{h}$

    $\displaystyle = \frac{ \frac{1}{(x + h)^3} - \frac{1}{x^3}}{h}$

    Multiply the numerator and denominator of the overall fraction by the LCD of $\displaystyle (x + h)^3$ and $\displaystyle x^3$:
    $\displaystyle = \frac{ \frac{1}{(x + h)^3} - \frac{1}{x^3}}{h} \cdot \frac{x^3(x + h)^3}{x^3(x + h)^3}$

    $\displaystyle = \frac{x^3 - (x + h)^3}{hx^3(x + h)^3}$

    Now expand the $\displaystyle (x + h)^3$ in the numerator and go from there.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nathan02079 View Post
    $\displaystyle \sqrt{1+(x^3-\frac{1}{4x^3})^2} $
    $\displaystyle \sqrt{1+ \left ( x^3-\frac{1}{4x^3} \right )^2} $

    $\displaystyle = \sqrt{1+ \left ( x^6 - 2 \frac{x^3}{4x^3} + \frac{1}{16x^6} \right )} $

    $\displaystyle = \sqrt{x^6 + 2 \frac{x^3}{4x^3} + \frac{1}{16x^6}} $
    In this last line I worked out $\displaystyle 1 - 2\frac{x^3}{4x^3} = \frac{4x^3}{4x^3} - \frac{2x^3}{4x^3} = \frac{2x^3}{4x^3}$ instead of simply making it $\displaystyle \frac{1}{2}$. This is so I can factor in the next step.

    So
    $\displaystyle \sqrt{1+ \left ( x^3-\frac{1}{4x^3} \right )^2} = \sqrt{x^6 + 2 \frac{x^3}{4x^3} + \frac{1}{16x^6}} $

    $\displaystyle = \sqrt{\left ( x^3 + \frac{1}{4x^3} \right )^2} = \pm \left ( x^3 + \frac{1}{4x^3} \right )$

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    $\displaystyle \frac{ (x + h)^{-3} - x^{-3}}{h}$

    $\displaystyle = \frac{ \frac{1}{(x + h)^3} - \frac{1}{x^3}}{h}$

    Multiply the numerator and denominator of the overall fraction by the LCD of $\displaystyle (x + h)^3$ and $\displaystyle x^3$:
    $\displaystyle = \frac{ \frac{1}{(x + h)^3} - \frac{1}{x^3}}{h} \cdot \frac{x^3(x + h)^3}{x^3(x + h)^3}$

    $\displaystyle = \frac{x^3 - (x + h)^3}{hx^3(x + h)^3}$

    Now expand the $\displaystyle (x + h)^3$ in the numerator and go from there.

    -Dan
    like this?

    $\displaystyle x^3-(x+h)(x+h)(x+h)\over hx^3(x+h)^3$

    $\displaystyle x^3-(x^3+2hx^2+xh^2+hx^2+2h^2x+h^3)\over hx^3(x+h)^3$

    $\displaystyle x^3-(x^3+3hx^2+3h^2x+h^3)\over hx^3(x+h)^3$

    $\displaystyle -3hx^2-3h^2x+h^3\over hx^3(x+h)^3$

    am i going in the right direction here?
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  5. #5
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    am i going in the right direction here?
    Looking a-okay to me!
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  6. #6
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by topsquark View Post
    $\displaystyle = \sqrt{\left ( x^3 + \frac{1}{4x^3} \right )^2} = \left| x^3 + \frac{1}{4x^3} \right|$

    -Dan
    Maybe?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post

    $\displaystyle = \sqrt{\left ( x^3 + \frac{1}{4x^3} \right )^2} = \left| x^3 + \frac{1}{4x^3} \right |
    $


    Maybe?
    Oh, probably. I always forget when to use principle value and when not to for some reason. (Which is embarrassing since I used to teach this. )

    -Dan
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