It's easy. If g commutes with A, then g commutes with A^3, because gAAA = AgAA = AAgA = AAAg.
If g commutes with A^3, then it commutes with A^6 = A^3 A^3, but A^6 = A, so it commutes with A.
Therefore, g commutes with A^3 iff it commutes with A, so C(A) = C(A^3).
However, the argument does not work when three divides the order of the element.
Take, for example, rotation by an angle that generates all other rotations as A, in D_6, the dihedral group of order 12; in other words, the symmetries of a regular hexagon. Then A^3 commutes with every element of the group, but A does not commute with reflections.
If a group has a finite integer order, it is not necessarily cyclic. The quaternion group and the dihedral groups are not cyclic; nor are the symmetry groups.
If a group is of prime order, though, then it *is* cyclic.