(Inspired by Indian Regional Mathematical Olympiad 2000, P1, untreated)

Find all pairs of positive integers $(x, y)$ such that $x^{3}+4 x^{2}-3 x+6=y^{3}$.
We have $(x+1)^{3}=x^{3}+3 x^{2}+3 x+1$ and $(x+2)^{3}=x^{3}+6 x^{2}+12 x+8$.

We can already state that for $x \geq 0, x^{3}+4 x^{2}-3 x+6<x^{3}+6 x^{2}+12 x+8=(x+2)^{3}$.

Furthermore, $x^{3}+4 x^{2}-3 x+6 \leq(x+1)^{3}$ if and only if $x^{2}-6 x+5 \leq 0$, if and only if $(x-1)(x-5) \leq 0$, if and only if $1 \leq x \leq 5$.

In other words, for $x=0$ or $x \geq 6$, we have $(x+1)^{3}<x^{3}+4 x^{2}-3 x+6<(x+2)^{3}$, so these are not solutions.

It remains to test $1 \leq x \leq 5$:

| $x$ | $x^{3}+4 x^{2}-3 x+6$ | Perfect cube? |
| :---: | :---: | :---: |
| 1 | $8=2^{3}$ | YES |
| 2 | 24 | no |
| 3 | 60 | no |
| 4 | 122 | no |
| 5 | $216=6^{3}$ | YES |

Finally, there are exactly two solutions, $(x, y)=(1,2)$ and $(x, y)=(5,6)$.