Irska

Let $ab+bc+ca=-p$ and $abc=q$. From $a+b+c=0$ we obtain $$0=(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2-2p,\text{ i.e.,}$$ $$a^2+b^2+c^2=2p,$$ and $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$, i.e., $$a^3+b^3+c^3=3q.$$ Also $$p^2=(ab+bc+ca{)}^2=a^2{b}^2+b^2{c}^2+c^2{a}^2+2abc(a+b+c)=a^2{b}^2+b^2{c}^2+c^2{a}^2,$$ and so $$4p^2=(a^2+b^2+c^2{)}^2=a^4+b^4+c^4+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)=a^4+b^4+c^4+2p^2,$$ whence $$a^4+b^4+c^4=2p^2.$$ Hence $$ $$\begin{array}{rl}(a-b{)}^2(b-c{)}^2(c-a{)}^2& =\det \left(\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right)\cdot \det \left(\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right)\\ & =\det \left(\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right)\cdot \left(\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right)\\ & =\det \left(\begin{array}{ccc}3& 0& 2p\\ 0& 2p& 3q\\ 2p& 3q& 2p^2\end{array}\right)\\ & =3(4p^3-9q^2)-8p^3\\ & =4p^3-27q^2\\ & =\frac{1}{2}(a^2+b^2+c^2{)}^3-27a^2{b}^2{c}^2.\end{array}$$