Mongolian Mathematical Olympiad

Answer: $F=-1/3,8/3$. Let us denote $s=a+b+c$. Then, under the given condition, we have $\frac{s}{c}=\frac{s}{a}=\frac{s}{b}$. (1) If $s=0$, then $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=3abc\ne 0$. If $s\ne 0$, then $a=b=c$. Hence $a^3+b^3+c^3=3a^3\ne 0$. This completes the proof of the first part.

(2) Since $a^3+b^3+c^3\ne 0$, the function $$F(a,b,c)=\frac{(a+b)(b+c)(c+a)}{a^3+b^3+c^3}$$ is well defined. Given $F(1,1,-2)=-1/3$ and $F(1,1,1)=8/3$, the values $-\frac{1}{3}$ and $\frac{8}{3}$ are attained because $1+1-2=0$ and $1+1+1\ne 0$. Now we show there are no other values of $F$. Indeed, if $s=0$, then $F=\frac{(-c)(-a)(-b)}{3abc}=-\frac{1}{3}$. If $s\ne 0$, then $a=b=c$, thus $F=\frac{8a^3}{3a^3}=\frac{8}{3}$.