Let $a,$ $b,$ $c$ be complex numbers such that
\begin{align*}
\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} &= 0, \\
\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} &= 1, \\
\frac{a^3}{b + c} + \frac{b^3}{c + a} + \frac{c^3}{a + b} &= 2.
\end{align*}
Find $abc.$
Let $s = a + b + c,$ and let $p = ab + bc + ca.$
We can write the given equations as
\begin{align*}
\frac{a}{s - a} + \frac{b}{s - b} + \frac{c}{s - c} &= 0, \\
\frac{a^2}{s - a} + \frac{b^2}{s - b} + \frac{c^2}{s - c} &= 1, \\
\frac{a^3}{s - a} + \frac{b^3}{s - b} + \frac{c^3}{s - c} &= 2.
\end{align*}We can write the first equation as
\[\frac{(s - a)(s - b)(s - c)}{(s - a)(s - b)} \cdot \frac{a}{s - a} + \frac{(s - a)(s - b)(s - c)}{(s - b)(s - c)} \cdot \frac{b}{s - b} + \frac{(s - a)(s - b)(s - c)}{(s - c)(s - a)} \cdot \frac{c}{s - c} = 0.\]Expanding, we get
\[a(s - a)(s - c) + b(s - a)(s - b) + c(s - b)(s - c) = 0.\]This simplifies to $s^3 - (a + b + c) s^2 + (ab + bc + ca) s - abc = 0.$
Now, note that
\begin{align*}
(ab + bc + ca)((a + b + c)^3 - a^3 - b^3 - c^3) &= (ab + bc + ca)(3s^2 (a + b + c) - 3s(a + b + c)^2 + 3s^3) \\
&\quad - (a^4 b + a^4 c + b^4 a + b^4 c + c^4 a + c^4 b) \\
&= 3p (p - 3s^2) + a^2 b^2 (a + b) + b^2 c^2 (b + c) + c^2 a^2 (c + a).
\end{align*}By Vieta's formulas, $s = \frac{-p}{1}$ and $p = \frac{2}{1} = 2,$ so
\begin{align*}
(ab + bc + ca)((a + b + c)^3 - a^3 - b^3 - c^3) &= 6 (-2)(-2^2) + a^2 b^2 (a + b) + b^2 c^2 (b + c) + c^2 a^2 (c + a) \\
&= 6 (8) + a^2 b^2 (a + b) + b^2 c^2 (b + c) + c^2 a^2 (c + a) \\
&= 48 + a^2 b^2 (a + b) + b^2 c^2 (b + c) + c^2 a^2 (c + a).
\end{align*}On the other hand,
\begin{align*}
(ab + bc + ca)((a + b + c)^3 - a^3 - b^3 - c^3) &= ((ab + bc + ca)(3s^2) - (ab + bc + ca)(3s) + (ab + bc + ca)(3s)) - (a^4 b + a^4 c + b^4 a + b^4 c + c^4 a + c^4 b + a^3 b^2 + a^3 c^2 + b^3 a^2 + b^3 c^2 + c^3 a^2 + c^3 b^2) \\
&= 48 + (ab + bc + ca)(3s) - a^3 b^2 - b^3 c^2 - b^2 a^3 - c^3 a^2 - c^2 b^3 - a^2 c^3 \\
&\quad - (a^4 b + a^4 c + b^4 a + b^4 c + c^4 a + c^4 b + a^3 b^2 + a^3 c^2 + b^3 a^2 + b^3 c^2 + c^3 a^2 + c^3 b^2).
\end{align*}Therefore,
\[48 + a^2 b^2 (a + b) + b^2 c^2 (b + c) + c^2 a^2 (c + a) = 48 + (ab + bc + ca)(3s) - a^4 b - a^4 c - b^4 a - b^4 c - c^4 a - c^4 b - a^3 b^2 - a^3 c^2 - b^3 a^2 - b^3 c^2 - c^3 a^2 - c^3 b^2.\]From the given equations,
\[2 = a^3 + b^3 + c^3 = (a + b + c)((a + b + c)^2 - 3(ab + bc + ca)) = 3s^3 - 9p,\]so
\[s^3 = \frac{2 + 9 \cdot 2}{3} = 9.\]Then from the equation $s^3 - (a + b + c) s^2 + (ab + bc + ca) s - abc = 0,$
\[9 - 9s + 2 = abc.\]Then from the equation $\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b},$
\[\frac{s^2 - 2p}{s} + \frac{8p}{s} = 1,\]or $s^2 - 2p + 8p = s,$ or $9 - 2(2) + 8 \cdot 2 = s = 21.$ Hence, $9 - 9 \cdot 21 + 2 = \boxed{-180}.$