#1
Precal... dont answer if you are a dumbass.

(x^6) - 1 .... factor completely
i broke it into (x^3 - 1)(x^3 + 1) using the difference of 2 squares
then into (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1) using difference of 2 cubes twice...
now i am stuck... i dont think it is factored as far as it can go, is it?
thanks
#3
Quote by rage__against10
4.6
its called parythmitic dumass.


How is that factoring it?
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#10
It can't go furthur man. Id have to sit down and work it out to make sure its correct, but that bottom line can't be factored furthur
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