2018/017) Solve for non -ve integers x, y $y^3=x^3+8x^2-6x+8$

We have
$y^3-x^3= 8x^2-6x+ 8>=8$ for $x>=0$
And $(x+3)^3 = x^3 + 9x^2 + 27x + 27 >y^3$
So we need to consider $y=x+1$ and  and $y=x+2$
Putting $y=x+1$ we get $x^3+3x^2+ 3x + 1 = x^3+8x^2-6x+8$
Or $5x^2-9x+7=0$
This equation does not have any real solution
Putting $y=x+2$ we get $x^3+6x^2+ 12x + 8 = x^3+8x^2-6x+8$
Or $2x^2-18x=0$
$=>x(x-9)=0$
Giving x = 0 or 9
That is solution set $(0,2)$ and $(9,11)$