Find all integers $x, y \in \mathbb{N}$ such that $y^{6}+2 y^{3}-y^{2}+1=x^{3}$.
We notice that for $y \geq 0$, we have $y^{4} \geq y^{3} \geq y^{2}$, thus

$$
\left(y^{2}\right)^{3}0$, then $0=3 y^{2}-2 y+4=2 y^{2}+(y-1)^{2}+3>0$, which is absurd.

Thus, $y=0$ and we verify that $(1,0)$ indeed provides a solution.

Alternative solution: Consider a potential pair of solutions $(x, y) \in \mathbb{N}^{2}$. Then, by factoring, we have: $\left(y^{3}+y+1\right)\left(y^{3}-y+1\right)=x^{3}$. Now, $\left(y^{3}+y+1\right) \wedge\left(y^{3}-y+1\right)=\left(y^{3}+y+1\right) \wedge 2 y=$ $\left(y^{3}+y+1\right) \wedge y=1$ because $y^{3}+y+1$ is odd.

Thus, $y^{3}+y+1=a^{3}$ and $y^{3}-y+1$ are both cubes. However, $y^{3}<y^{3}+y+1 \leqslant(y+1)^{3}$. The inequality is an equality only when $y=0$, in which case $(x, y)=(1,0)$. Otherwise: $y^{3}<$ $a^{3}<(y+1)^{3}$, which is absurd!

The only solution is therefore $(x, y)=(1,0)$.