What is the fundamental theorem of algebra for x^6-64?

Answer 1

I would understand it to say that
x^6 - 64 = 0 has 6 roots

Proof:
first treat x^6 as a difference of squares
x^6 - 64
= (x^3 - 8)(x^3 + 8)
now as a difference of cubes for the first factor, and a sum of cubes for the second part
= (x-2)(x^2 + 2x + 4)(x+2)(x^2 - 2x + 4)

so x^6 - 64 = 0
(x-2)(x^2 + 2x + 4)(x+2)(x^2 - 2x + 4) = 0

x-2=0 ----> x=2
x+2=0 ----> x=-2
x^2+2x+4)=0 ---> x = (-2 ± √-12)/2 = -1 ± √3 i
x^2 - 2x + 4=0 --> x = (2 ± √-12)/2 = 1 ± √3 i

6 roots!