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:

Write the expression inside the tenth power as

$$\Delta = \frac{x+1}{x^{2/3}+1-x^{1/3}}-\frac{x+1}{x-x^{1/2}},\qquad x\gt 1$$

Step 1 (first fraction)
Factor the denominator as $$1-x^{1/3}+x^{2/3}=\dfrac{1+x}{1+x^{1/3}}$$ because
$$1-x^{1/3}+x^{2/3}=(x^{1/3})^{2}-x^{1/3}+1=\frac{x+1}{x^{1/3}+1}.$$
Hence

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

Step 2 (second fraction)
Factor $$x$$ from the denominator:

$$\frac{x+1}{x-x^{1/2}} =\frac{x+1}{x\bigl(1-x^{-1/2}\bigr)} =\frac{1+x^{-1}}{1-x^{-1/2}}.$$ For $$x\gt 1$$ we have $$|x^{-1/2}|<1,$$ so use the geometric series $$\frac{1}{1-x^{-1/2}}=\sum_{k=0}^{\infty}x^{-k/2}.$$ Thus

$$\frac{x+1}{x-x^{1/2}} =(1+x^{-1})\sum_{k=0}^{\infty}x^{-k/2} =1+x^{-1/2}+2x^{-1}+2x^{-3/2}+2x^{-2}+\cdots\;.$$

Step 3 (series for Δ)
Subtract the series just obtained from $$1+x^{1/3}:$$

$$\Delta=(1+x^{1/3})- \bigl[1+x^{-1/2}+2x^{-1}+2x^{-3/2}+2x^{-2}+\cdots\bigr] =x^{1/3}-x^{-1/2}-2x^{-1}-2x^{-3/2}-2x^{-2}-\cdots\;.$$

Keep the powers of $$x$$ together with their coefficients:

Term T1: coefficient $$1$$, power $$x^{1/3}$$.
Term T2: coefficient $$-1$$, power $$x^{-1/2}$$.
Term T3: coefficient $$-2$$, power $$x^{-1}$$.
Term T4: coefficient $$-2$$, power $$x^{-3/2}$$, and so on.

Step 4 (condition for constant term in $$\Delta^{10}$$)
Choose $$a$$ copies of T1, $$b$$ copies of T2, $$c$$ copies of T3, $$d$$ copies of T4, … in the multinomial expansion, where

$$a+b+c+d+\cdots = 10\qquad -(1)$$ and the total exponent of $$x$$ must be zero:

$$\frac{1}{3}a-\frac{1}{2}b-1c-\frac{3}{2}d-\cdots = 0.$$

Multiply by $$6$$ to avoid fractions:

$$2a-3b-6c-9d-\cdots = 0\qquad -(2)$$

Because every negative coefficient in $$(2)$$ is a multiple of $$3,$$ the left-hand side can vanish only if $$2a$$ is also a multiple of $$3.$$
Hence $$a$$ itself must be a multiple of $$3,$$ so $$a=0,3,6,9$$ (but $$\le 10$$).

Case a = 0: the sum in $$(2)$$ is negative ⇒ impossible.

Case a = 3: $$2a=6.$$ Then $$(2)\Rightarrow 3b+6c+9d+\cdots = 6 \Rightarrow b+2c+3d+\cdots = 2.$$ Equation $$(1)$$ gives $$b+c+d+\cdots = 7,$$ impossible together.

Case a = 6: $$2a=12.$$ Then $$b+2c+3d+\cdots = 4\qquad -(3)$$ and $$b+c+d+\cdots = 4\qquad -(4).$$ Subtract $$(4)$$ from $$(3)$$ to get $$c+2d+3e+\cdots = 0,$$ which forces $$c=d=e=\cdots = 0.$$ Consequently $$b=4.$$ Thus the only viable solution is
$$a=6,\; b=4,\; c=d=e=\cdots =0.$$

Case a = 9: leads to contradictions similar to the case $$a=3$$ (no solution).

Therefore the constant term arises solely from choosing
6 factors of $$x^{1/3}$$ and 4 factors of $$-x^{-1/2}.$$

Step 5 (coefficient)
The multinomial coefficient is $$\dfrac{10!}{6!4!} = 210.$$ The numerical factors from the chosen terms are $$(1)^6(-1)^4 = 1.$$

Hence the term independent of $$x$$ in $$\Delta^{10}$$ is $$210.$$

Answer : Option A  - 210