Mark M. answered 10/12/21
Retired math prof. Very extensive Precalculus tutoring experience.
By "solution set", I assume that you mean: the values of x for which f(x) = 0
x4 - 256 = 0
(x2)2 - (16)2 = 0
(x2 - 16)(x2 + 16) = 0
(x + 4)(x - 4)(x2 + 16) = 0
x = 4, -4, or x2 = -16
x = 4, -4, 4i, -4i (each solution has multiplicity one)