Proof for rationality of a number

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.

• from where did you got this problem? Commented Sep 16, 2018 at 14:07
• It’s from Central America Mathematical Olympiad (I’m not certain if it’s the formal name).
– user542970
Commented Sep 16, 2018 at 14:09
• I suspect that in order for the first sum to be rational, each term must be rational. For example, if two of the terms were rational but the third wasn't, the sum would not be rational. If the cube root of a quantity is rational, then the cube root of its inverse is also rational, hence the second sum would also be rational. Commented Sep 16, 2018 at 17:43

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 $x = \sqrt[3]{ab^2}$, $y = \sqrt[3]{bc^2}$, $z = \sqrt[3]{ca^2}$.