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Math Help - Simplifying

  1. #1
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    Simplifying

    Hello,

    Not sure if I've simplified this as far as I can - pretty sure I can go further, just not sure what I should be doing (factoring, distributive)

    Equation is (z + 1)^2 (3xy^2)^3 / (9y^2)^(1/2) x^(-3) (z+1)^(-1)

    I've simplified it to...

    (z^2 + 1) (27x^3y^6) / 3y (1/x^3) (1/z+1)

    What should I be looking at next? Finding a common factor between both numerator and denominator (I can see three, along with each of the variables). I'd assume before I can factor out though, I should be expanding the brackets further - is that possible at this stage, and if so, how do I go about doing the denominator.

    As you can see, I'm pretty lost as to what I should be doing next, so any tips would be great!
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  2. #2
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    Re: Simplifying

    Quote Originally Posted by astuart View Post
    Hello,

    Not sure if I've simplified this as far as I can - pretty sure I can go further, just not sure what I should be doing (factoring, distributive)

    Equation is (z + 1)^2 (3xy^2)^3 / (9y^2)^(1/2) x^(-3) (z+1)^(-1)

    I've simplified it to...

    (z^2 + 1) (27x^3y^6) / 3y (1/x^3) (1/z+1)

    What should I be looking at next? Finding a common factor between both numerator and denominator (I can see three, along with each of the variables). I'd assume before I can factor out though, I should be expanding the brackets further - is that possible at this stage, and if so, how do I go about doing the denominator.

    As you can see, I'm pretty lost as to what I should be doing next, so any tips would be great!
    \displaystyle \begin{align*} (z + 1)^2 \neq z^2 + 1 \end{align*}...
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  3. #3
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    Re: Simplifying

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} (z + 1)^2 \neq z^2 + 1 \end{align*}...
    Gah, )z+1)^2 would equal z^2 + 2z + 1 or something along those lines, right?

    I think i'll tackle this in the morning, not at 1 in the morning.
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  4. #4
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    Re: Simplifying

    Also \displaystyle \begin{align*} (z + 1)^{-1} \neq \frac{1}{z} + 1 \end{align*}, it's \displaystyle \begin{align*} \frac{1}{z + 1} \end{align*}
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  5. #5
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    Re: Simplifying

    Quote Originally Posted by Prove It View Post
    Also \displaystyle \begin{align*} (z + 1)^{-1} \neq \frac{1}{z} + 1 \end{align*}, it's \displaystyle \begin{align*} \frac{1}{z + 1} \end{align*}
    Right, I thought that was the case, but I just wasn't sure how to use the equation tags properly.

    So I have...
    (z^2+2z+1) (9x^3y^6) / (3y)(1/x^3)(1/Z+1)

    Not sure if I an simplify between the numerator and denominator. There's a common factor of three there. Can I cancel out the z and the x? I didn't think I could because the z and the x are underneath a fraction (As in underneath a 1).
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