Originally Posted by

**melese** I try a straighforward approach.

The integers $\displaystyle a,b,c$ cannot be equal; otherwise $\displaystyle 3a=a+b+c=abc=a^3$ and then $\displaystyle a^2=3$, but $\displaystyle 3$ is not a perfect square.

So the following inequality holds: $\displaystyle a+b+c<c+c+c=3c$. Now if $\displaystyle ab\geq3$, then $\displaystyle 3c\leq abc$ implying $\displaystyle a+b+c<abc$.

Therefore, $\displaystyle ab<3$, i.e. $\displaystyle a=1$ and $\displaystyle b=2$. The equation $\displaystyle a+b+c=abc$ gives $\displaystyle 3+c=2c$ and hence $\displaystyle c=3$.