The term independent of $$x$$ in the expansion of $$\left[\frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x - x^{1/2}}\right]^{10}$$, $$x \neq 1$$, is equal to ___.

We first simplify the expression inside the brackets. For the first fraction, note that $$x + 1 = (x^{1/3})^3 + 1 = (x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)$$, so $$\dfrac{x+1}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$$.

For the second fraction, factor the denominator: $$x - x^{1/2} = x^{1/2}(x^{1/2} - 1)$$, and the numerator: $$x - 1 = (x^{1/2} - 1)(x^{1/2} + 1)$$. So $$\dfrac{x-1}{x - x^{1/2}} = \dfrac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)} = \dfrac{x^{1/2}+1}{x^{1/2}} = 1 + x^{-1/2}$$.

Subtracting: $$(x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$$. We need the term independent of $$x$$ in $$(x^{1/3} - x^{-1/2})^{10}$$.

The general term is $$\binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r = \binom{10}{r}(-1)^r\,x^{(10-r)/3 - r/2}$$. Setting the exponent to zero: $$\dfrac{10-r}{3} - \dfrac{r}{2} = 0$$, which gives $$2(10-r) = 3r$$, so $$r = 4$$.

The term independent of $$x$$ is $$\binom{10}{4}(-1)^4 = 210$$.