1. Simplify the expression inside the bracket

We need the constant term in $\left( \frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}} \right)^{10}.$

Let $t=x^{1/3}.$ Then $x=t^3,\quad x^{2/3}=t^2,\quad x^{1/2}=t^{3/2}.$ But a cleaner route is to simplify each fraction directly.

2. First fraction

Use the identity $(a+1)(a^2-a+1)=a^3+1.$

Taking $a=x^{1/3}$, we get $\bigl(x^{1/3}+1\bigr)\bigl(x^{2/3}-x^{1/3}+1\bigr)=x+1.$

Hence $\frac{x+1}{x^{2/3}-x^{1/3}+1}=x^{1/3}+1.$

3. Second fraction

Factor the denominator: $x-x^{1/2}=x^{1/2}(x^{1/2}-1).$ Also, $x-1=(x^{1/2}-1)(x^{1/2}+1).$

Therefore, \frac{x-1}{x-x^{1/2}}=rac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)}=\frac{x^{1/2}+1}{x^{1/2}}=1+x^{-1/2}.

4. Expression inside the bracket

So the quantity becomes $\left[(x^{1/3}+1)-(1+x^{-1/2})\right]^{10}=(x^{1/3}-x^{-1/2})^{10}.$

Factor out $x^{-1/2}$: $x^{1/3}-x^{-1/2}=x^{-1/2}\left(x^{5/6}-1\right).$

Thus $(x^{1/3}-x^{-1/2})^{10}=x^{-5}(x^{5/6}-1)^{10}.$

5. Find the constant term

Expand: $x^{-5}(x^{5/6}-1)^{10}=x^{-5}\sum_{k=0}^{10}\binom{10}{k}(x^{5/6})^k(-1)^{10-k}.$

General term is $\binom{10}{k}(-1)^{10-k}x^{-5+\frac{5k}{6}}.$

For the term independent of $x$, exponent must be zero: $-5+\frac{5k}{6}=0.$ Multiply by $6$: $-30+5k=0\implies k=6.$

So the constant term is $\binom{10}{6}(-1)^{4}=\binom{10}{6}=210.$

6. Check options

• A: $4$ ❌
• B: $120$ ❌
• C: $210$ ✅
• D: $310$ ❌

Therefore, the required term independent of $x$ is $\boxed{210}.$