Finding rational solutions to $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but $abc\notin\mathbb{Z}$ I'm trying to find $a,b,c\in\mathbb{Q}$ such that $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but that $abc\notin\mathbb{Z}$.
I tried to pick specific values for $a, b$
$$
b^3x^2 + a^3x - n= 0
$$
where I tried different integers $n\neq 0$ and then evaluate the roots of this polynomial. However, this proved unsuccessful and now I'm stuck.

COMMENT.-Certainly you do want to have $a,b,c\in\mathbb Q\setminus\mathbb Z$ so I am afraid you have affaire to very particular elliptic curves of the form $Ax^3+BY^3=CZ^3$ which have been seriously studied first by E. S. Selmer (The diophantine equation $ax^3+by^3+cz^3=0$ Act. Math. (Stockh.) 85 (1951) p.203-362). It is a topic in which the difficulty is very hard mainly if you want to impose restrictive conditions like yours.
In fact, put $\dfrac{a}{\alpha}, \dfrac{b}{\beta},\dfrac{c}{\gamma}$ your rational with $\dfrac{abc}{\alpha\beta\gamma}\notin\mathbb Z$ then your condition becomes
$$\gamma c(\beta a)^3+c^2(\alpha b)^3=n(\alpha\gamma)^3$$
However you have at hand some divisibility conditions you can maybe used to get divers class of solutions. In any case you do have to try with a Selmer curve.
