The number of positive integral values taken by $f(x)=(x^3-11^3)^{1/3}$ over natural numbers $x$ 
For $f:{\Bbb{N}}\to \mathbb R$, $f(x)=(x^3-11^3)^{1/3}$, what is the number of positive integral values $f(x)$ can take?

The first thing that came into my mind was just simply trial and error but I soon realised that it wasn't going to come handy here.   I thought maybe factoring
$$(x^3-11^3)= (x-11)(x^2 + 11x + 121)$$
might give some leads to this but I just couldn't figure something out from this.
How do I approach this? I believe there might be some kind of specific theorem or rule that helps to solve these kind of problems that I don't know of, because I just can't figure out how to start in the first place.
 A: The answer is $0$ by Fermat's Last Theorem. Indeed, $(x^3-11^3)^{1/3}=y \in \mathbb{N}$ would imply $y^3+11^3=x^3$, which has no solutions in the positive integers by FLT.
A: Indeed, the Fermat's Last Theorem  comes to mind. However, with the hint from @lulu, we can show that the equation
$$y^3+ 11^3 = x^3$$
has no positive solutions. Otherwise, we had
$$11^3 = (x-y)(x^2+ x y + y^2)$$ and so $x-y$ is a power of $11$. If it were at least $11$, then $x^2 + x y + y^2$ must equal $11^2$, so $y=0$. Otherwise, we have $x-y=1$, so $x^3-y^3= 3y^2 + 3 y + 1\not = 11^3$ (check $\mod 3$).
A: x must be 11^3 * (k^3+1), with a non-negative integer constant k, in order that f(x) will become an integer too. But x then will become 11 * (k^3+1)^(1/3). This itself will only be an integer as required, when k = 0, with f(11) = 0. This though is not a positive integral value.
=> The number of positive integer values f may take is zero.
P.S.: Third roots of negative numbers are excluded.
A: If nothing else, there is always one of The Oldest Tricks In The Book: The difference between any other cube and $x^3$ is at least $3x^2-3x+1$, which is greater than $11^3$ for any $x \ge 25$. So if nothing else one can try all integers between 1 and 25.
