Solution:
We observe that
Q=a3+b3+c3−3abc=21(a+b+c)((a−b)2+(b−c)2+(c−a)2)
Since we are looking for the least positive value taken by Q, it follows that a,b,c are not all equal. Thus a+b+c≥1+1+2=4 and (a−b)2+(b−c)2+(c−a)2≥1+1+0=2. Thus we see that Q≥4. Taking a=1, b=1 and c=2, we get Q=4. Therefore the least value of Q is 4 and this is achieved only by a+b+c=4 and (a−b)2+(b−c)2+(c−a)2=2. The triples for which Q=4 are therefore given by
(a,b,c)=(1,1,2), (1,2,1), (2,1,1)