Dividing Rational Expressions

x^3-8/x^2-4 divided by (x^2+2x+4)/(x^3+8). This becomes ((x-2)(x^2+2x+4)/(x+2)(x-2) times ((x+2)(x^2-2x+4))/(x^2+2x+4). I understand that (x-2)(x^2+2x+4)=(x^3-8) when I distribute it, however, if I didn't have the help of my book, I'd have no clue how to come up with this on my own. Is there a general rule for how to distribute x's to the third power? Thank you!!

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asked by Jaz
  1. yes!
    We can factor both the sum and the difference of cubes

    A^3 + b^3 = (A+B)(A^2 - AB + B^2) and
    A^3 - b^3 = (A-B)(A^2 + AB + B^2)

    that is where those factors came from.
    Most of the stuff canceled and you should have had x^2 - 2x + 4 as the answer.

    posted by Reiny

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