TITLE: Find $x\in \mathbb{Z}$ such that $54x^3+1$ is a cube
QUESTION [11 upvotes]: Find $x\in \mathbb{Z}$ such that $54x^3+1$ is a cube.

I found $x=0$, any others ?

REPLY [4 votes]: The curve $54x^3+1=y^3$ is the equation of an elliptic curve. Let $C: 54U^3+V^3=W^3$ be the projectivization of the original curve, and let us declare $[0,1,1]$ to be the origin for the addition law on the elliptic curve.
The change of variables 
$$\begin{cases} U = X/18,\\ V=-Y,\\ W=-Y+Z/324,\\
\end{cases}$$
gives a birational equivalence of $C$ and $E: X^3 - Y^2Z + 1/324YZ^2 - 1/314928 Z^3=0$, which we can de-homogenize to be $E: y^2 - 1/324 y = x^3 - 1/314928$ (which in turn can be further simplified to be $y^2=x^3-27$, but we will not use this last model here). Using the method of descent, one can show that the rank of $E$ is zero, and the torsion subgroup is of order $2$. Hence, there are only two rational points on $E$, namely $[0,1,0]$ and $[1/108,1/648,1]$. These correspond to the only two rational points on the curve $C$, namely
$$(0,1) \quad \text{ and } \quad (-1/3,-1).$$
In particular, the only rational numbers $x\in\mathbb{Q}$ such that $54x^3+1$ is also a cube are $x=0$ and $x=-1/3$.