1. Simplify the expression inside the bracket

We need the constant term in

$\left(\frac{x+1}{x^{2/3}+1-x^{1/3}}-\frac{x-1}{x-x^{1/2}}\right)^{10}, \qquad x>1.$

Let

$t=x^{1/3}.$ Then $x=t^3$ and

$x^{2/3}+1-x^{1/3}=t^2+1-t=t^2-t+1.$

Also,

$x+1=t^3+1=(t+1)(t^2-t+1).$

Hence,

$\frac{x+1}{x^{2/3}+1-x^{1/3}}=\frac{(t+1)(t^2-t+1)}{t^2-t+1}=t+1=x^{1/3}+1.$

Now simplify the second fraction:

$\frac{x-1}{x-x^{1/2}}=\frac{x-1}{x(1-x^{-1/2})}.$

A better factorization is:

$x-1=(\sqrt{x}-1)(\sqrt{x}+1),$ $x-x^{1/2}=\sqrt{x}(\sqrt{x}-1).$

So,

$\frac{x-1}{x-x^{1/2}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}=\frac{\sqrt{x}+1}{\sqrt{x}}=1+x^{-1/2}.$

Therefore the bracket becomes

$\left(x^{1/3}+1-(1+x^{-1/2})\right)^{10}=(x^{1/3}-x^{-1/2})^{10}.$

2. Rewrite for binomial expansion

$(x^{1/3}-x^{-1/2})^{10}.$

General term is

$T_{r+1}=\binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r$

$=\binom{10}{r}(-1)^r x^{\frac{10-r}{3}-\frac r2}.$

Power of $x$ is

$\frac{10-r}{3}-\frac r2=\frac{20-2r-3r}{6}=\frac{20-5r}{6}.$

For the term independent of $x$, set exponent $=0$:

$\frac{20-5r}{6}=0$ $20-5r=0$ $r=4.$

3. Find the constant term

For $r=4$,

$T_5=\binom{10}{4}(-1)^4=\binom{10}{4}=210.$

So the term independent of $x$ is

$\boxed{210}.$

4. Compare with stored answer

Stored correct answer: D

Our derived answer: D (210)

They agree.