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    (Original post by thegreatwhale)
    lol still don't know where the 3 comes from and where the 8x^3 -1 went
     \displaystyle \frac{8x^3-1}{1-2x^3} \equiv \frac{-4(1-2x^3)+3}{1-2x^3} .
    You can separate this into 2 separate fractions  \displaystyle \frac{-4(1-2x^3)}{1-2x^3}+\frac{3}{1-2x^3} = -4+\frac{3}{1-2x^3} .
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    (Original post by thegreatwhale)
    woah where did that come from?? o..o
    i understand \dfrac{1}{1-2x^3} =(1-2x^3)^{-1}

    but i'm a little unsure of where the 3 comes from ad i'm assuming the -4 is to balance things out
    \dfrac{8x^3-1}{1-2x^3}=\dfrac{-4(1-2x^3)+3}{1-2x^3}=-4 \times \dfrac{1-2x^3}{1-2x^3}+\dfrac{3}{1-2x^3}=3(1-2x^3)^{-1}-4

    It's not necessary. I noticed there there was an x^3 in the numerator and denominator and no other powers of x so I simplified and differentiated at the time. I wouldn't have used polynomial long division for a calculus problem if it were more complicated.
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    (Original post by fefssdf)
    I'm actually so confused right now
    me 2
    (Original post by B_9710)
     \displaystyle \frac{8x^3-1}{1-2x^3} \equiv \frac{-4(1-2x^3)+3}{1-2x^3} .
    You can separate this into 2 separate fractions  \displaystyle \frac{-4(1-2x^3)}{1-2x^3}+\frac{3}{1-2x^3} = -4+\frac{3}{1-2x^3} .
    (Original post by Kvothe the arcane)
    \dfrac{8x^3-1}{1-2x^3}=\dfrac{-4(1-2x^3)+3}{1-2x^3}=-4 \times \dfrac{1-2x^3}{1-2x^3}+\dfrac{3}{1-2x^3}=3(1-2x^3)^{-1}-4
    ah awesome ^-^ for got you could "take out a common factor from anything as long as u balance it" thanks
 
 
 
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