The term independent of $$x$$ in the expansion of $$\left(\frac{(x+1)}{\left(x^{2/3} + 1 - x^{1/3}\right)} - \frac{(x-1)}{\left(x - x^{1/2}\right)}\right)^{10}$$, $$x > 1$$ is:

Simplifying the first rational expression using the algebraic identity $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
where $$a = x^{1/3}$$ and $$b = 1$$:
$$x + 1 = \left(x^{1/3}\right)^3 + 1^3 = \left(x^{1/3} + 1\right)\left(x^{2/3} - x^{1/3} + 1\right)$$

Dividing by the denominator:
$$\frac{x+1}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$$

Simplifying the second rational expression by factoring both the numerator and denominator:
$$\frac{x-1}{x - x^{1/2}} = \frac{\left(x^{1/2} - 1\right)\left(x^{1/2} + 1\right)}{x^{1/2}\left(x^{1/2} - 1\right)} = \frac{x^{1/2} + 1}{x^{1/2}} = 1 + x^{-1/2}$$

Subtracting the two simplified expressions:
$$\left(x^{1/3} + 1\right) - \left(1 + x^{-1/2}\right) = x^{1/3} - x^{-1/2}$$

The given expression simplifies to:
$$\left(x^{1/3} - x^{-1/2}\right)^{10}$$

Writing the general term $$T_{r+1}$$ in the binomial expansion:
$$T_{r+1} = \binom{10}{r} \left(x^{1/3}\right)^{10-r} \left(-x^{-1/2}\right)^r$$
$$T_{r+1} = (-1)^r \binom{10}{r} x^{\frac{10-r}{3} - \frac{r}{2}}$$

For the term independent of $$x$$, setting the exponent of $$x$$ equal to $$0$$:
$$\frac{10-r}{3} - \frac{r}{2} = 0$$
$$2(10 - r) - 3r = 0$$
$$20 - 5r = 0 \implies r = 4$$

Substituting $$r = 4$$ back into the term:
$$T_5 = (-1)^4 \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$

Hence, the term independent of $$x$$ is $$210$$.