Ireland_2017

So, suppose $a$, $b$, $c$ satisfy the given equations, and eliminate $c$, say. Then, from the first, we deduce that $$\begin{array}{rl}{a}^3+b^3+c^3& =a^3+b^3-(a+b{)}^3\\ & =a^3+b^3-(a^3+3a^{2}b+3a{b}^2+b^3)\\ & =-3ab(a+b)\\ & =3abc.\end{array}$$ 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.