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 $p_n/q_n$.
Best Answer
This is a very hard proof to do for an undergraduate but there are books available. Tthe book "Invitation to Fermat Wiles" (http://www.amazon.com/Invitation-Mathematics-Fermat-Wiles-Yves-Hellegouarch/dp/0123392519) is an exposition on the proof written for undergraduates for example.