
2. Determine the least positive value taken by the expression $a^{3}+b^{3}+c^{3}-3 a b c$ as $a, b, c$ vary over all positive integers. Find also all triples $(a, b, c)$ for which this least value is attained.


Solution: We observe that

$$
\left.Q=a^{3}+b^{3}+c^{3}-3 a b c=\frac{1}{2}(a+b+c)\right)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)
$$

Since we are looking for the least positive value taken by $Q$, it follows that $a, b, c$ are not all equal. Thus $a+b+c \geq 1+1+2=4$ and $(a-b)^{2}+(b-$ $c)^{2}+(c-a)^{2} \geq 1+1+0=2$. Thus we see that $Q \geq 4$. 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

$$
(a, b, c)=(1,1,2),(1,2,1),(2,1,1)
$$
