Harvard-MIT Mathematics Tournament

Solution:
Let $s=a+b+c$. Then
$$\begin{aligned} s^{3} & =a^{3}+b^{3}+c^{3}+3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right)+6 a b c \\ & =a b c\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}+3\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right)+6\right) \\ & =a b c\left(1+\left(3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9\right)+6\right) \\ & =a b c\left(3 s\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-2\right) \\ & =a b c(3 s \bar{s}-2) \quad(\text{ because } \bar{s}=\bar{a}+\bar{b}+\bar{c}=1 / a+1 / b+1 / c) \\ & =a b c\left(3|s|^{2}-2\right) \end{aligned}$$
Taking absolute values, we find $|s|^{3}=\left|3|s|^{2}-2\right|$. It follows that $|s|$ must be a positive real root of $x^{3}-3 x^{2}+2=0$ or $x^{3}+3 x^{2}-2=0$. However, since the negative real roots of $x^{3}-3 x^{2}+2=0$ are exactly the additive inverses of the positive real roots of $x^{3}-3 x^{2}+2=0$, and all three roots of $x^{3}-3 x^{2}+2=0$ are real $(x^{3}-3 x^{2}+2=0$ may be factored as $(x-1)(x^{2}-2 x-2)=0$, and the discriminant of $x^{2}-2 x-2$ is positive), the product of all possible values of $|s|$ is $(-2) \cdot(-1)^{n}$, where $n$ denotes the number of negative real roots of $x^{3}-3 x^{2}+2=0$. By Descartes's Rule of Signs, we see that $n$ is odd, so the answer is 2, as desired.