1. Since $c = -a - b$, we get:
$$\begin{aligned} 0 &= ab^3 + b(-a-b)^3 + (-a-b)a^3 \\ &= -(ab^3 + b(a+b)^3 + (a+b)a^3) \\ &= -(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4) \\ &= -(a^2(a+b)^2 + b^2(a+b)^2 + a^2b^2) \\ &= -a^2c^2 - b^2c^2 - a^2b^2. \end{aligned}$$
Therefore each term of the last sum must be zero: $ab = bc = ca = 0$. Hence two of the numbers must be zero and from the equality $a + b + c = 0$, we conclude that $a = b = c = 0$.