Let $a, b, c\neq 0$ be rational numbers such that $\sqrt[3]{ab^{2}}+\sqrt[3]{bc^{2}}+\sqrt[3]{ca^{2}}\neq 0$ is a rational number. Prove that $$\sqrt[3]{\frac{1}{ab^{2}}}+\sqrt[3]{\frac{1}{bc^{2}}}+\sqrt[3]{\frac{1}{ca^{2}}}$$ is also rational.
Note. This is a number theory problem, thus I am asking for a number theoretic proof.
I started by denoting $x=\sqrt[3]{ab^{2}}+\sqrt[3]{bc^{2}}+\sqrt[3]{ca^{2}}$ and $y=\sqrt[3]{\frac{1}{ab^{2}}}+\sqrt[3]{\frac{1}{bc^{2}}}+\sqrt[3]{\frac{1}{ca^{2}}}$ and tried proving that numbers like $xy$ or $\frac{x}{y}$ are rational, a thus $y$ is also rational, but I couldn’t complete my proof.
(Ugh... The point of departure was here. Still it is purely algebraic.)
For nonzero $x, y, z$ and an integer $n$, let $P_n = x^n + y^n + z^n$. Then $$3xyz(1 - P_1 P_{-1}) = P_3 - P_1^3$$ (this can be verified directly). It remains to put $x = \sqrt[3]{ab^2}$, $y = \sqrt[3]{bc^2}$, $z = \sqrt[3]{ca^2}$.