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Thread: More simplifying help.

  1. #1
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    More simplifying help.

    Can someone show me how

    $\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}$

    is te he same as



    $\displaystyle \frac{x^3}{(1-xy)^2}-3x^3$

    They are apparently alternative ways to write each other but i have no idea how!
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  2. #2
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    Hello el123
    Quote Originally Posted by el123 View Post
    Can someone show me how

    $\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}$

    is te he same as



    $\displaystyle \frac{x^3}{(1-xy)^2}-3x^3$

    They are apparently alternative ways to write each other but i have no idea how!
    Sorry, but it just isn't.

    You can factorise the numerator:

    $\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}=\frac{x^3(4-3xy)}{(1-xy)^2}$

    and that's all.

    Grandad
    Last edited by Grandad; Sep 17th 2009 at 12:20 AM.
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  3. #3
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    that $\displaystyle 3x^3 $
    is supposed to be

    $\displaystyle \frac{3x^3}{xy-1}$

    check it right here.

    - Wolfram|Alpha
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  4. #4
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    Hello el123
    Quote Originally Posted by el123 View Post
    that $\displaystyle 3x^3 $
    is supposed to be

    $\displaystyle \frac{3x^3}{xy-1}$

    check it right here.

    - Wolfram|Alpha
    Yes, it's true you can write $\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}$ as $\displaystyle \frac{x^3}{(1-xy)^2}-\frac{3x^3}{xy-1}$ if you need to! How does it work? Well, note first that $\displaystyle (1-xy)^2 = (xy-1)^2$, and then multiply top and bottom of the second fraction by (xy-1):

    $\displaystyle \frac{x^3}{(1-xy)^2}-\frac{3x^3}{xy-1}$

    $\displaystyle =\frac{x^3}{(1-xy)^2}-\frac{3x^3}{(xy-1)}\color{red}\frac{(xy-1)}{(xy-1)}$

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

    $\displaystyle =\frac{4x^3-3x^4y}{(1-xy)^2}$

    Grandad
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