Cubic polynomial equal to a cube I've been researching cubes and I'm trying to solve this Diophantine equation over the integers.
$$ax^3 + bx^2 + cx + d = y^3$$where a, b, c, d are parameters for a given $n$. For example, for $n = 5$, you have$$5x^3 + 30x^2 + 90x + 100 = y^3$$
Is there any way to solve this Diophantine equation? Or does anyone know any references or links to point me to? (Either for general $a,b,c,d$ or for the specific one above). Thanks!
Also a, b, c, and d are functions of n:
$$a = n$$
$$b = (3/2)n(n-1)$$
$$c = (1/2)n(2n-1)(n-1)$$
$$d = (1/4)n^2(n-1)^2$$
It's theorized that there are no solutions (to this specific one). But I want to prove it. Surely, I can't check for every single x, I'm mainly looking for a bound on x as a function of n. Essentially, "After some finite x, it will never be a cube" ...
 A: As you pointed out in a comment, you transformed the equation into $y^3=Ax^3+Bx=x(Ax^2+B)$. From that it is easy to show that any prime factor of $Ax^2+B$ which is not a prime factor of $AB$, does not divide $x$, so because of the equation it shows up with multiplicity $3$. The upshot is that you can write $Ax^2+B = Cz^3$, where $C$ is not unique but is constrained to a finite set. Now multiply this equation by $A^3C^2$ and get $V^2 = U^3 + K, V = A^2Cx, U = ACy, K =-BA^3C^2$. This is a Mordell equation and Sage and Magma will solve it, there many papers with explicit bounds, pretty much everything you want to know. 
A: There is a short section about equations like yours in the book History of the Theory of Numbers, Volume II, by Leonard Eugene Dickson (page 566). It deals with rational solutions not integer ones, but it might be helpful/interesting for you if you haven't already seen it.
I think I have found a counterexample to: "It's theorized that there are no solutions (to this specific one)."
I tried the case $n = 8$ on a hunch (since 8 is a cube). This gives $a = 8$, $b = 84$, $c = 420$, $d = 784$, and thus gives rise to the equation
$8x^3 + 84x^2 + 420x + 784 = y^3$
Setting $y = x + 7$ gives the integer solution $x = -3$.
