# anyone smart here? Prime Factoring quick question please

x^6-64

I'm thinking its (x^3-8)(x^3+8) from a difference of cubes, but from a difference of squares it can be (x^2-4)(x^4+4x^2+16).

But then neither of those are prime...

can anyone help me out?
Larrivée D-03R
US Fender Telecaster, Vox AC15
heres what you do

the first thing you had right with the difference of squares actually
take it to (x^3-8)*(x^3+8)

after that, there are formulas for sum and difference of cubes, i think. use them to factor the remaining equations. you can do it separately then throw it all back together if it makes it that much easier

btw, 2^3=8

hope that helped
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i got (x+2)(x-2)(x+2)(x-2)(x+2)(x-2)

would that be the prime polynomials?

Also, I just heard a program I have on my laptop, Maple 11, which can solve x^6-64 for me, and gave me (x-2)(x+2)(x^2 + 2x + 4)(x^2 - 2x + 4)

would mine still be right though? I just don't know how Maple 11 got that one, doesn't show the work xO lol
Larrivée D-03R
US Fender Telecaster, Vox AC15