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

We begin with the given expression

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

Our target is the term that does not contain $$x$$ after expanding the entire power.

First we simplify each fraction inside the large brackets.

For the first fraction, we put $$t=x^{1/3} \;(\text{so}\; x=t^{3})$$. Then

$$x^{2/3}=t^{2},\qquad x^{1/3}=t.$$

Hence

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

Now we recall the algebraic identity $$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2}).$$ Setting $$a=t,\;b=1$$ gives

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

Substituting into the numerator, we get

$$\dfrac{(t+1)(t^{2}-t+1)}{t^{2}-t+1}=t+1.$$

Undoing the substitution $$t=x^{1/3},$$ this simplifies to

$$x^{1/3}+1.$$

Next we simplify the second fraction. Put $$s=x^{1/2}\;(\text{so}\;x=s^{2}).$$ Then

$$x=s^{2},\qquad x^{1/2}=s.$$

Thus

$$\dfrac{x-1}{x-x^{1/2}}=\dfrac{s^{2}-1}{s^{2}-s}.$$

Factorising numerator and denominator, we use $$a^{2}-b^{2}=(a-b)(a+b)$$ to write

$$s^{2}-1=(s-1)(s+1),\qquad s^{2}-s=s(s-1).$$

Hence

$$\dfrac{(s-1)(s+1)}{s(s-1)}=\dfrac{s+1}{s}=1+\dfrac{1}{s}=1+s^{-1}.$$

Replacing $$s=x^{1/2}$$ gives

$$1+x^{-1/2}.$$

So the entire bracket now becomes

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

Therefore the whole expression is

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

To expand, we use the Binomial Theorem, which states

$$ (u+v)^{n}=\sum_{k=0}^{n}\binom{n}{k}u^{\,n-k}v^{\,k}. $$

Here $$u=x^{1/3}$$ and $$v=-x^{-1/2},$$ with $$n=10.$$ The general term is

$$T_{k}=\binom{10}{k}(x^{1/3})^{\,10-k}\,(-x^{-1/2})^{\,k}.$$

Simplify the powers of $$x$$ in this term. The exponent of $$x$$ in $$T_{k}$$ is

$$\left(10-k\right)\!\left(\dfrac13\right)+k\!\left(-\dfrac12\right) =\dfrac{10-k}{3}-\dfrac{k}{2}.$$

For the term to be independent of $$x,$$ this exponent must be zero:

$$\dfrac{10-k}{3}-\dfrac{k}{2}=0.$$

Multiply by the common denominator $$6$$:

$$2(10-k)-3k=0.$$

So

$$20-2k-3k=0\quad\Longrightarrow\quad 20-5k=0\quad\Longrightarrow\quad k=4.$$

We substitute $$k=4$$ back into the general term:

$$T_{4}=\binom{10}{4}(x^{1/3})^{\,10-4}\,(-x^{-1/2})^{\,4} =\binom{10}{4}(x^{1/3})^{\,6}\,(-1)^{4}(x^{-1/2})^{\,4}.$$

Compute each factor:

$$(x^{1/3})^{6}=x^{6/3}=x^{2},\qquad (x^{-1/2})^{4}=x^{-2},\qquad (-1)^{4}=1.$$

The product $$x^{2}\cdot x^{-2}=x^{0}=1,$$ so all powers of $$x$$ cancel as required.

The coefficient is therefore

$$\binom{10}{4}=210.$$

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

Hence, the correct answer is Option A.