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 $a^3+b^3+c^3 = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc = 3abc \neq 0$. If $s \neq 0$, then $a = b = c$. Hence $a^3 + b^3 + c^3 = 3a^3 \neq 0$. This completes the proof of the first part.

(2) Since $a^3 + b^3 + c^3 \neq 0$, 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 $1+1+1 \neq 0$. 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 $s \neq 0$, then $a = b = c$, thus $F = \frac{8a^3}{3a^3} = \frac{8}{3}$.