The factors of 120 are (1,120), (2,60), (3,40), (4,30), (5,24), (6,20), (8,15), (10,12). There are no plus or minus signs so one of these factors must apply. Out of these factors we have two perfect cubes: 1=1^3 and 8=2^3. Since 1 isn't present and cannot be achieved by division, we're left with the last quantity 2^3. That leaves 12 and 2, which cannot be combined to form 15 so that 15*8=120. Starting at the other end with 12, we would need to find 10 in the other numbers, and the only combination I can find is 12*(2+2^3)=120, but that requires us to change two operators from / and * to * and +. So there doesn't seem to be a solution with the given operators. Anyone?