Key Concept:

To find the term independent of $x$ in the expansion of a complex expression raised to a power, we follow a two-step approach:

1. Simplify the base expression: Use algebraic identities and exponent rules to transform the given expression into a standard binomial form, typically $(Ax^p + Bx^q)^n$. This simplification is crucial as it reduces the complexity of subsequent calculations.
2. Apply the Binomial Theorem: Use the general term formula for binomial expansion, $T_{r+1} = \binom{n}{r} (first~term)^{n-r} (second~term)^r$, to find the term where the power of $x$ is zero.

---

Step-by-Step Solution:

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

1. Simplify the first part of the expression: $\frac{x+1}{x^{2/3} - x^{1/3} + 1}$

• Objective: Simplify this fraction by recognizing algebraic identities.
• Strategy: Observe that the numerator $x+1$ can be seen as a sum of cubes. Let $a = x^{1/3}$ and $b = 1$. Then $x = (x^{1/3})^3$ and $1 = 1^3$.
• Applying the Sum of Cubes Identity: Recall the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$. Substituting $a = x^{1/3}$ and $b = 1$: $x+1 = (x^{1/3})^3 + 1^3 = (x^{1/3}+1)((x^{1/3})^2 - x^{1/3} \cdot 1 + 1^2)$ $x+1 = (x^{1/3}+1)(x^{2/3} - x^{1/3} + 1)$
• Performing the Cancellation: Now substitute this back into the fraction: $\frac{x+1}{x^{2/3} - x^{1/3} + 1} = \frac{(x^{1/3}+1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1}$ Since $x>1$, the denominator $x^{2/3} - x^{1/3} + 1$ is never zero, allowing us to cancel the common factor. $= x^{1/3} + 1$
  • Explanation: We intentionally rewrote the numerator as a sum of cubes because its factored form contained the exact expression present in the denominator. This allowed for direct cancellation, leading to a significantly simpler term.

2. Simplify the second part of the expression: $\frac{x-1}{x-x^{1/2}}$

• Objective: Simplify this fraction by recognizing algebraic identities and factoring.
• Strategy (Numerator): The numerator $x-1$ can be seen as a difference of squares. Let $a = \sqrt{x}$ (or $x^{1/2}$) and $b = 1$. Then $x = (\sqrt{x})^2$ and $1 = 1^2$.
• Applying the Difference of Squares Identity: Recall the identity $a^2-b^2 = (a-b)(a+b)$. Substituting $a = \sqrt{x}$ and $b = 1$: $x-1 = (\sqrt{x})^2 - 1^2 = (\sqrt{x}-1)(\sqrt{x}+1)$
• Strategy (Denominator): The denominator $x-x^{1/2}$ has a common factor of $x^{1/2}$ (or $\sqrt{x}$).
• Factoring the Denominator: $x - x^{1/2} = x^{1/2}(x^{1/2} - 1) = \sqrt{x}(\sqrt{x}-1)$
• Performing the Cancellation: Now substitute these back into the fraction: $\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}$ Since $x>1$, $\sqrt{x}-1$ is not zero, allowing us to cancel the common factor. $= \frac{\sqrt{x}+1}{\sqrt{x}}$
• Further Simplification: Divide each term in the numerator by the denominator: $= \frac{\sqrt{x}}{\sqrt{x}} + \frac{1}{\sqrt{x}} = 1 + \frac{1}{x^{1/2}}$ To prepare for binomial expansion, it's often useful to express terms with negative exponents: $= 1 + x^{-1/2}$
  • Explanation: We factored both the numerator using the difference of squares and the denominator by extracting a common term. This strategic factorization allowed for cancellation, simplifying the second part of the expression into a more manageable form.

3. Combine the simplified parts into a single binomial expression

• Objective: Substitute the simplified forms of the first and second terms back into the original expression and simplify further.
• Substitution: $\left( (x^{1/3}+1) - \left(1 + x^{-1/2}\right) \right)^{10}$
• Distribution of Negative Sign and Simplification: $\left( x^{1/3}+1 - 1 - x^{-1/2} \right)^{10}$ The $+1$ and $-1$ cancel out: $\left( x^{1/3} - x^{-1/2} \right)^{10}$
  • Explanation: By combining the highly simplified terms, the entire complex base expression is reduced to a simple binomial form $(a+b)^n$. This is the crucial step that enables the application of the binomial theorem.

4. Find the general term of the binomial expansion

• Objective: Use the binomial theorem to find a formula for any term in the expansion.
• Identifying Components: Our expression is $(x^{1/3} - x^{-1/2})^{10}$. Here, $n=10$, $A=x^{1/3}$, and $B=-x^{-1/2}$.
• Applying the General Term Formula: The $(r+1)$-th term ($T_{r+1}$) in the expansion of $(A+B)^n$ is given by: $T_{r+1} = \binom{n}{r} A^{n-r} B^r$ Substitute the values: $T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r$
• Separating Coefficients and Powers of x: $T_{r+1} = \binom{10}{r} x^{\frac{1}{3}(10-r)} (-1)^r (x^{-1/2})^r$ $T_{r+1} = \binom{10}{r} (-1)^r x^{\frac{10-r}{3}} x^{-\frac{r}{2}}$
• Combining Powers of x: Using the rule $x^a \cdot x^b = x^{a+b}$: $T_{r+1} = \binom{10}{r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$
  • Explanation: The general term formula allows us to express any term in the expansion in a structured way. By carefully combining the powers of $x$, we obtain an expression for the exponent of $x$ that depends only on $r$, which is essential for the next step.

5. Determine the value of $r$ for the term independent of $x$

• Objective: Find the specific value of $r$ for which the term does not contain $x$.
• Condition for Term Independent of x: For a term to be independent of $x$, the exponent of $x$ must be $0$.
• Setting the Exponent to Zero: $\frac{10-r}{3} - \frac{r}{2} = 0$
• Solving for r: Find a common denominator (6) for the fractions: $\frac{2(10-r)}{6} - \frac{3r}{6} = 0$ Multiply both sides by 6 to clear the denominators: $2(10-r) - 3r = 0$ $20 - 2r - 3r = 0$ $20 - 5r = 0$ $5r = 20$ $r = 4$
  • Explanation: By setting the exponent of $x$ to zero, we are mathematically defining the condition for the term to be independent of $x$. The resulting linear equation in $r$ is then solved to find the specific index ($r$) that yields this term. The value $r=4$ is a valid integer between $0$ and $n=10$.

6. Calculate the term independent of $x$

• Objective: Substitute the found value of $r$ back into the general term formula to get the constant term.
• Substituting r=4: The term independent of $x$ is $T_{r+1} = T_{4+1} = T_5$. $T_5 = \binom{10}{4} (-1)^4 x^{\frac{10-4}{3} - \frac{4}{2}}$ We know that $(-1)^4 = 1$ and the exponent of $x$ will be $0$ (making $x^0 = 1$). $T_5 = \binom{10}{4} (1) (1)$ $T_5 = \binom{10}{4}$
• Calculating the Binomial Coefficient: $\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$ $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4 \times 3 \times 2 \times 1 \times 6!}$ Cancel out $6!$: $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$ $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{24}$ $\binom{10}{4} = 10 \times \frac{9}{3} \times \frac{8}{4 \times 2} \times 7 = 10 \times 3 \times 1 \times 7$ $\binom{10}{4} = 210$
  • Explanation: With $r$ determined, we directly calculated the binomial coefficient $\binom{n}{r}$, which gives us the numerical value of the specific term. Since the powers of $x$ cancel out to $x^0=1$, the coefficient itself is the term independent of $x$.

---

Tips and Common Mistakes to Avoid:

• Prioritize Simplification: Always spend adequate time simplifying the base expression. Most errors in these problems stem from incorrect initial algebraic manipulations rather than the binomial theorem itself.
• Exponent Rules: Be meticulous with fractional and negative exponents. A common mistake is misapplying rules like $(x^a)^b = x^{ab}$ or $x^a \cdot x^b = x^{a+b}$.
• Sign Management: Pay close attention to negative signs within the binomial. If the second term is negative (like $-x^{-1/2}$), remember to include $(-1)^r$ in the general term.
• Validate 'r': After calculating $r$, ensure it is a non-negative integer and $0 \le r \le n$. If $r$ is a fraction or negative, it indicates an error in calculation or that such a term does not exist in the expansion.
• Binomial Coefficient Calculation: Double-check the calculation of $\binom{n}{r}$. It's often easier to cancel terms in the factorial expansion before multiplying large numbers.

---

Summary and Key Takeaway:

This problem is an excellent example of how algebraic simplification (using identities like sum of cubes and difference of squares) combined with the binomial theorem can solve complex expressions. The critical steps involve reducing the given expression to a standard binomial form $(Ax^p + Bx^q)^n$, then using the general term $T_{r+1}$ to extract the power of $x$, and finally setting that power to zero to find the index $r$ of the term independent of $x$. Mastering algebraic identities and exponent rules is fundamental to efficiently tackling such problems.

The final answer is $\boxed{\text{210}}$.