Functionally Odd

The denominator of $f(x)$ can be written as

$\large (x + 1)^{\frac{2}{3}} + ((x + 1)(x - 1))^{\frac{1}{3}} + (x - 1)^{\frac{2}{3}} = \frac{(x + 1) - (x - 1)}{(x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}} = \dfrac{2}{(x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}}$ ,

where we made use of the identity $a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$ . Thus the desired sum is

$\large \dfrac{1}{2} \displaystyle\sum_{k=1}^{500} (((2k - 1) + 1)^{\frac{1}{3}} - ((2k - 1) - 1)^{\frac{1}{3}}) = \dfrac{1}{2}\left(\sum_{k=1}^{500} (2k)^{\frac{1}{3}} - \sum_{k=1}^{500} (2(k - 1))^{\frac{1}{3}}\right) =$

$\large \displaystyle \dfrac{1}{2} \left(\sum_{k=1}^{500} (2k)^{\frac{1}{3}} - \sum_{n=0}^{499} (2n)^{\frac{1}{3}}\right) = \dfrac{1}{2}((2*500)^{\frac{1}{3}} - (2*0)^{\frac{1}{3}}) = \dfrac{1000^{\frac{1}{3}}}{2} = \dfrac{10}{2} = \boxed{5}.$

Rationalize the denominator using difference of cube formula. Then the summation becomes a telescoping sequence.