TITLE: "Rationalizing" an equation
QUESTION [5 upvotes]: $$x=\sqrt[3]{p}+\sqrt[3]{q}$$
I'm trying to figure out some way to "rationalize" the previous equation, meaning to rewrite it purely in terms of whole number powers of $p$, $q$, and $x$. It seems quite simple but I've been stuck trying to do it. I'd appreciate anyone's help with this.

REPLY [7 votes]: Following up on Bernard's suggestion:
$$x= \sqrt[3]{p}+\sqrt[3]{q}$$
$$x(\sqrt[3]{p^2}-\sqrt[3]{pq}+\sqrt[3]{q^2})= p+q$$
$$x((\sqrt[3]{p}+\sqrt[3]{q})^2-3\sqrt[3]{pq})= p+q$$
$$x(x^2-3\sqrt[3]{pq})= p+q$$
$$x^3-3x\sqrt[3]{pq}= p+q$$
$$3x\sqrt[3]{pq}= x^3-p-q$$
$$27x^3pq= (x^3-p-q)^3$$