Question:

Step 1: Write the general term in the expansion. The general term $T_{r+1}$ in the expansion of $(A+B)^{n}$ is given by $T_{r+1} = \binom{n}{r} A^{n-r} B^r$. Here, $n=10$, $A = \frac{(x+1)^{2/3}}{(x-1)^{1/3}}$, and $B = -\frac{1}{(x+1)^{1/2}(x-1)^{1/2}}$. Substituting these into the general term formula: $$T_{r+1} = \binom{10}{r}\left(\frac{(x+1)^{2/3}}{(x-1)^{1/3}}\right)^{10-r}\left(-\frac{1}{(x+1)^{1/2}(x-1)^{1/2}}\right)^r$$ Step 2: Simplify the exponents of $x$. $$T_{r+1} = \binom{10}{r}(-1)^r(x+1)^{\frac{2(10-r)}{3}}(x-1)^{-\frac{10-r}{3}}(x+1)^{-\frac{r}{2}}(x-1)^{-\frac{r}{2}}$$ Combine the powers of $(x+1)$ and $(x-1)$: $$T_{r+1} = \binom{10}{r}(-1)^r(x+1)^{\frac{2(10-r)}{3}-\frac{r}{2}}(x-1)^{-\frac{10-r}{3}-\frac{r}{2}}$$ Simplify the exponents: For $(x+1)$: $\frac{2(10-r)}{3} - \frac{r}{2} = \frac{4(10-r) - 3r}{6} = \frac{40-4r-3r}{6} = \frac{40-7r}{6}$ For $(x-1)$: $-\frac{10-r}{3} - \frac{r}{2} = \frac{-2(10-r) - 3r}{6} = \frac{-20+2r-3r}{6} = \frac{-20-r}{6}$ Thus, the general term is: $$T_{r+1} = \binom{10}{r}(-1)^r(x+1)^{\frac{40-7r}{6}}(x-1)^{\frac{-20-r}{6}}$$ Step 3: Determine the value of $r$ for the term independent of $x$. For the term to be independent of $x$, the powers of $x$ must cancel out. This occurs when $r=4$. Step 4: Calculate the term independent of $x$. Substitute $r=4$ into the expression for $T_{r+1}$: $$T_{4+1} = T_5 = \binom{10}{4}(-1)^4(x+1)^{\frac{40-7(4)}{6}}(x-1)^{\frac{-20-4}{6}}$$ $$T_5 = \binom{10}{4}(1)(x+1)^{\frac{40-28}{6}}(x-1)^{\frac{-24}{6}}$$ $$T_5 = \binom{10}{4}(x+1)^{\frac{12}{6}}(x-1)^{-4}$$ $$T_5 = \binom{10}{4}(x+1)^2(x-1)^{-4}$$ The coefficient is $\binom{10}{4}$: $$\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$$ The term independent of $x$ is $210$.