So, suppose $a$, $b$, $c$ satisfy the given equations, and eliminate $c$, say. Then, from the first, we deduce that
$$\begin{aligned} a^3 + b^3 + c^3 &= a^3 + b^3 - (a + b)^3 \\ &= a^3 + b^3 - (a^3 + 3a^2b + 3ab^2 + b^3) \\ &= -3ab(a + b) \\ &= 3abc. \end{aligned}$$
This and the third equation forces $abc = 0$. Hence, one of $a$, $b$, $c$ is zero. Say $c = 0$. Then, by the first and second equations, $a = -b$, and $1 = 2a^2$. Thus one solution is $a = \pm 1/\sqrt{2}$, $b = \mp 1/\sqrt{2}$, $c = 0$, and any permutation of this triple is a solution. Conversely, every such triple is a solution.