Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
            $$\implies y^3-x^3=8x^2-6x+8$$
            $$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the right track?Please help.

Since we have$$y^3-x^3=2(4x^2-3x+4)$$
there exists an integer $k$ such that
$$y-x=2k\iff y=x+2k.$$
So, we have
$$(x+2k)^3-x^3=2(4x^2-3x+4)\iff (3 k-4) x^2+(6 k^2+3) x+4k^3-4=0\tag1$$
Now we have
$$D=(6k^2+3)^2-4(3k-4)(4k^3-4)\ge 0\iff -12 k^4+64 k^3+36 k^2+48 k-55\ge 0$$$$\iff 12 k^4-64 k^3-36 k^2-48 k+55\le 0$$
Here, let $f(k)=12 k^4-64 k^3-36 k^2-48 k+55=4k[k\{k(3k-16)-9\}-12]$.
Now, for $k\ge 6$, we have
$$3k-16\ge 2\Rightarrow k(3k-16)\ge 2k\ge 12$$$$\Rightarrow k\{k(3k-16)-9\}-12\ge 3k-12\gt 0\Rightarrow f(k)\gt 0.$$
Also, for $k\le -1$, we have
$$f(k)=12 k^4-36k^2(k+1) -48 k+55-28k^3\gt 0.$$
Hence, we have $k=0,1,2,3,4,5.$
Then, from $(1)$, you can find integer roots $x$ for each $k$.
