how do you factor 4x^3-108

Kimberly,

in the first you indicate the sum, in the second you indicate a difference (minus). Since they are similar, we'll address both.

For the first, this also looks like the sum of two cubes (a^{3} + b^{3}) = (a + b) (a^{2} - ab + b^{2}). But 4 isn't a perfect cube, and neither is 108. BUT... if we factor out a 4, we get

4(x^{3} + 27) = 4 (x + 3)(x^{2} - 3x + 9)

For the second, we approach it the same way, except our signs change:

Difference of two cubes (a^{3} - b^{3}) = (a - b) (a^{2} + ab + b^{2}).

4(x^{3} - 27) = 4 (x - 3)(x^{2} - 3x + 9)