Find $x\in \mathbb{Z}$ such that $54x^3+1$ is a cube

Find $x\in \mathbb{Z}$ such that $54x^3+1$ is a cube.

I found $x=0$, any others ?

I think the trivial solution you got is the only solution. We need $54x^3 + 1 = y^3$. Let $v =3x$ and $u = y$. Then the equation is: $$1 - u^3 = 2v^3$$

However, there is an early result by Euler (I think) which says that the sum or difference of two cubes cannot be double another cube, unless it is the trivial case of both the cubes being the same.

EDIT: I notice someone had already commented on very similar lines. The Euler reference can be obtained here GoogleBook Theorem 247.

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 $x=-1/3$.