By Fermat's last theorem, the equation $u^3+v^3=w^3$ has no solutions
in positive integers $u,v,w$. Now consider the following variant : call $\rho(x)$
the distance between $x$ and the nearest integer, for any real number $x$
(thus $\rho(3)=0,\rho(3.2)=0.2,\rho(3.5)=0.5, \rho(3.9)=0.1$ etc). 


An "approximate" version of the Fermat equation is to ask
for $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ to be arbitrarily small. A trivial
way to achieve this is to make one of the variables very small
compared to the other, say $v$ very small compared to $u$, so that
${( u^3+v^3 )}^{\frac{1}{3}}$ is very near to $u$.


It is therefore natural to ask if there is an absolute constant $C>0$
such that $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ can be made arbitrarily small
with $u,v$ positive and $u \leq C v, v \leq C u$ (so that neither of
$u$ or $v$ dominates).


Can a (reasonably) explicit $(u\_n,v\_n)$ sequence be found, such that
$\rho(u\_n^3+v\_n^3)$ tends to $0$ as $n$ goes to infinity and
$u\_n \leq C v\_n, v\_n \leq C u\_n$ ? "Closed-form" formula
would be the best, of course, but even a simple recursion would be nice.



Variant of Fermat's last theorem
The condition you want is very weak, and there are clearly many accidental solutions.


You can add severe restrictions and still find many solutions. For example, (as Steven Sivek pointed out) you can force $u\_n = v\_n$ and then by the theory of simple continued fractions, there are infinitely many $p\_n/q\_n$ so that 


$|\frac{p\_n}{q\_n}-\sqrt[3]2|\lt q\_n^{-2}.$ 


Then $|p\_n - q\_n \sqrt[3]2| < 1/q\_n$ so $\rho(\sqrt[3]{(q\_n^3+q\_n^3)} )$ decreases to 0 rapidly. 


This might not be viewed as explicit since the coefficients of the simple continued fraction for $2^{1/3}$ don't have a clear pattern, although you can define them recursively if you allow functions like $\lfloor1/\rho\rfloor$.


If you can find $x^3 + y^3 = z^3$ in numbers with known simple continued faction expansions, then you may be able to use this to construct closed form families of examples. For example, 


$(5-\sqrt{6})^3 + 3\sqrt{6}^3 = (5+\sqrt{6})^3$. 


Convergents of $\sqrt6$, $p\_n/q\_n$, satisfy $|\frac{p\_n}{q\_n} - \sqrt{6}| < 1/q\_n^2$. 


Then $(5q\_n-p\_n)^3+(3p\_n)^3 = (5q\_n+p\_n)^3 + O(q\_n) = (5q\_n+p\_n)^3 + o((5q\_n+p\_n)^2)$, 


so $\rho(\sqrt[3]{(5q\_n-p\_n)^3+(3p\_n)^3}) = o(1).$


For example, $\sqrt[3]{(5\times 881 - 2158)^3 + (3 \times 2158)^3} = 6552.99916...$


Since $\sqrt{6} = [2;2,4,2,4,2,4...]$ which is periodic, you can construct a closed form expression for the convergents $p\_n/q\_n$.


