Belarusian Mathematical Olympiad

Note that
$$\begin{align*} (ab - c^2)(ca - b^2)(bc - a^2) &= (a^2bc - ab^3 - ac^3 + b^2c^2)(bc - a^2) = \\ &= a^2b^2c^2 - ab^4c - abc^4 + b^3c^3 - a^4bc + a^3b^3 + a^3c^3 - a^2b^2c^2 = \\ &= a^3b^3 + b^3c^3 + a^3c^3 - a^4bc - a^4bc - abc^4 = a^3b^3 + b^3c^3 + a^3c^3 - abc(a^3 + b^3 + c^3). \end{align*}$$
Therefore the equality $a^3b^3 + b^3c^3 + c^3a^3 = abc(a^3 + b^3 + c^3)$ is equivalent to
$$(ab - c^2)(ca - b^2)(bc - a^2) = 0,$$
whence it follows that either $ab = c^2$ or $ca = b^2$ or $bc = a^2$, as required.