Polynomial Diophantine Equation 
If $x$,$y$ $\in \mathbb Z$, find all the solutions of
$$y^3=x^3+8x^2-6x+8$$

I have tried factorizing the equation but the polynomial on $\text{R.H.S.}$ doesn't have any integral roots. Further, I deduce that the parity of $x$ and $y$ is same. However, I can't seem to find a way to go further.
Any help will be appreciated.
Thanks!
 A: Suppose that $x\geq 0$.  Then we have $x^3 < x^3 + 8x^2 - 6x + 8 < (x+3)^3$, so $y \in \{x+1,x+2\}$.  $y=x+1$ has no integer solutions, and plugging in $y=x+2$ yields the two solutions $x=0$ and $x=9$.
For negative $x$, we can do something similar, though the details are messier:
Suppose $x\leq 0$, and let $u=-x$, $v=-y$.  Then $v^3 = u^3 - 8u^2 -6u - 8$.  We have $(u-12)^3<u^3 - 8u^2 -6u - 8<u^3$, so $v \in \{u-11, u-10, \ldots, u-1\}$.
Plugging in yields no integer solutions except for the ones we already found, $(x,y) \in\{(0,2),(9,11)\}$.
There may be a way to avoid difficulty in the last step, but the basic idea is to use bounds to reduce the problem to checking finitely many cases.
A: In general for these type of equations there are good upper bounds for $|x|.$ For instance see theorem 4 of a paper of G.Walsh [http://matwbn.icm.edu.pl/ksiazki/aa/aa62/aa6225.pdf].
This theorem gives $|x|\leq 3^9\cdot 9^6=3^{21}.$ Then you can check (having a little patience) all such $x$ (you need $2^{33}$ rounds and check only for $|x|<0$ due to the answer of the previous poster; it is doable).
