What techniques would be use to prove $6x^2 +12x +8$ cannot be perfect cube for integer x > 0 I'm wondering if there are any basic techniques.

Let $(x,y)$ be any integer solution to $y^3 = 6x^2 + 12x + 8$.
Since RHS is even, so does LHS and $y$. This implies $\text{RHS} \equiv 0 \pmod 8$.
It is easy to check
$$\text{LHS} \equiv 
\begin{cases} 0,& x \text{ even}\\2, & x \text{ odd}\end{cases}
\pmod 8$$
So $x$ is also even. This implies existence of integers $X,Y$ such that $x = 2X$ and $y = 2Y$.
In terms of $X, Y$, we have
$$Y^3 = 3X^2 + 3X + 1 = (X+1)^3 - X^3
\quad\iff\quad X^3 + Y^3 = (X+1)^3$$
By Fermat's Last theorem, at least one of $X, Y$ or $X+1$ is $0$.
This leads to three and only three set of possibilities:


*

*$X =  0 \rightarrow (x,y) = (0,2)$

*$Y =  0 \rightarrow \text{ no solution}$.

*$X = -1 \rightarrow (x,y) = (-2,2)$


Since all of these have $x \le 0$, $6x^2 + 12x + 8$ is never a cube for $x > 0$.
