Question

Did I solve this problem correctly?
Directions: State the possible rational zeros for each function.
Question:f(x) = x6 - 64
Answer:
Constant term:64 Factors:1,2,4,8,16,32,64
Leading coefficient:1 Factors: 1
±1,2,4,8,16,32,64/1
= ±1, ±2, ±4, ±8, ±16, ±32, ±64

Answer 1

Please use the same nickname throughout your different posts.

You are making things way too complicated.
You are looking at a difference of squares

x^6 - 64
= (x^3 -8)(x^3 + 8)
now I see a sum and a difference of cubes
= (x-2)(x^2 + 2x + 4)(x+2)(x^2 -2x + 4)

the two quadratics don't factor over the rationals so they will not yield rational zeros

x = 2 or x = -2 are the rational zeros

Answer 2

NO, NO , NO.

f(x)=0=x^6-64
x^6=64
take sixth root of each side,
x=+-2

Now there are other roots, but not real, I suspect you have not had DeMoirves theorem, but in polar comples plane
x^6=64@360deg
x=2@n60 where n=0,1,2,3,4,5, and the n=0 or 3 gives the +-2 real roosts, with no imaginary part.

Answer 3

As I read it, the directions were to list the possible rational zeros, not to actually find them.

Looks to me like Truda did it just right.