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
$$p^2 = (ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) = a^2b^2 + b^2c^2 + c^2a^2,$$
and so
$$4p^2 = (a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2) = a^4 + b^4 + c^4 + 2p^2,$$
whence
$$a^4 + b^4 + c^4 = 2p^2.$$
Hence
$$$$
\begin{align*}
(a-b)^2 (b-c)^2 (c-a)^2 &= \det \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix} \cdot \det \begin{pmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{pmatrix} \\
&= \det \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix} \cdot \begin{pmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{pmatrix} \\
&= \det \begin{pmatrix} 3 & 0 & 2p \\ 0 & 2p & 3q \\ 2p & 3q & 2p^2 \end{pmatrix} \\
&= 3(4p^3 - 9q^2) - 8p^3 \\
&= 4p^3 - 27q^2 \\
&= \frac{1}{2}(a^2 + b^2 + c^2)^3 - 27a^2b^2c^2.
\end{align*}