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!!

Question

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!!

Reiny · Answer

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.