The problem with fist factoring x^6-64 as a difference of cubes is that x^6-64=(x^2-4)(x^4+4x^2+16). One may notice that (x^2-4) factors as a difference of squares to (x+2)(x-2); however, the second factor does not fit any patterns. It does factor it, but using the a-c method, one is led to believe that (x^4+4x^2+16) is a prime polynomial. The only way to see that (x^4+4x^2+16) factors is to use the last two factors of the answer (x+2)(x-2)(x^2-2x+4)(x^2+2x+4). Multiply (x^2-2x+4)(x^2+2x+4) and the result should be (x^4+4x^2+16), but this is not a result you would have found using any pattern of factoring quadratics. If you first factor as a difference of squares, then as a difference of cubes, you'll factor it completely. If you first factor the difference of cubes, you will, most likely, not be able to finish the problem. The real catch to the problem is to see that it can be done either way and to experiment with how the results differ. One path leads to a more factored result than the other. Hope that helps.
