1. Key Concepts and Formulas

This problem requires a combination of algebraic simplification and the application of the Binomial Theorem.

• Binomial Theorem: For any positive integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^{n} {}^{n}C_r a^{n-r} b^r$ where ${}^{n}C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient. The $(r+1)^{th}$ term in the expansion is $T_{r+1} = {}^{n}C_r a^{n-r} b^r$. A term is "independent of $x$" if the exponent of $x$ in that term is zero.

• Algebraic Identities: We will utilize the following identities for simplification:

  • Sum of Cubes: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
  • Difference of Squares: $a^2 - b^2 = (a-b)(a+b)$

• Properties of Exponents: We'll use rules like $(x^m)^n = x^{mn}$ and $x^m \cdot x^n = x^{m+n}$.

2. Step-by-Step Solution

The problem asks for the term independent of $x$ in the expansion of ${\left[ {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right]^{10}}$. Our first step is to simplify the complex expression inside the large bracket.

Step 2.1: Simplify the first term inside the bracket The first term is ${{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}}$. We can rewrite $x+1$ using the sum of cubes identity. Let $a = x^{1/3}$ and $b = 1$. Then $x = (x^{1/3})^3$ and $1 = 1^3$. So, $x+1 = (x^{1/3})^3 + 1^3$. Applying the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$: $(x^{1/3})^3 + 1^3 = (x^{1/3} + 1)((x^{1/3})^2 - x^{1/3} \cdot 1 + 1^2) = (x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)$ Now, substitute this back into the first term: ${{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} = {{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)} \over {x^{2/3} - x^{1/3} + 1}}$ Since $x \ne 1$, the denominator $x^{2/3} - x^{1/3} + 1 \ne 0$, so we can cancel the common factor: $= x^{1/3} + 1$ Explanation: We recognized the numerator as a sum of cubes, which allowed us to factor it in a way that matches the denominator, leading to a significant simplification.

Step 2.2: Simplify the second term inside the bracket The second term is ${{x - 1} \over {x - {x^{1/2}}}}$. We can rewrite $x-1$ using the difference of squares identity. Let $a = \sqrt{x}$ and $b = 1$. Then $x = (\sqrt{x})^2$ and $1 = 1^2$. So, $x-1 = (\sqrt{x})^2 - 1^2$. Applying the identity $a^2 - b^2 = (a-b)(a+b)$: $(\sqrt{x})^2 - 1^2 = (\sqrt{x} - 1)(\sqrt{x} + 1)$ For the denominator $x - x^{1/2}$, we can factor out $x^{1/2}$ (which is $\sqrt{x}$): $x - x^{1/2} = x^{1/2}(x^{1/2} - 1) = \sqrt{x}(\sqrt{x} - 1)$ Now, substitute these back into the second term: ${{x - 1} \over {x - {x^{1/2}}}} = {{(\sqrt{x} - 1)(\sqrt{x} + 1)} \over {\sqrt{x}(\sqrt{x} - 1)}}$ Given $x \ne 1$, we know $\sqrt{x} \ne 1$, so $\sqrt{x} - 1 \ne 0$. Thus, we can cancel the common factor: $= {{\sqrt{x} + 1} \over {\sqrt{x}}}$ This can be further split into two terms: $= {\sqrt{x} \over {\sqrt{x}}} + {1 \over {\sqrt{x}}} = 1 + x^{-1/2}$ Explanation: We applied the difference of squares identity to the numerator and factored the denominator to find a common factor, simplifying the expression significantly.

Step 2.3: Combine the simplified terms Now substitute the simplified first and second terms back into the original expression: ${\left[ {(x^{1/3} + 1) - (1 + x^{-1/2})} \right]^{10}}$ Distribute the negative sign: ${\left[ {x^{1/3} + 1 - 1 - x^{-1/2}} \right]^{10}}$ The constant terms ($+1$ and $-1$) cancel out: ${\left[ {x^{1/3} - x^{-1/2}} \right]^{10}}$ Explanation: Careful handling of the subtraction is crucial here. The goal is to reduce the expression to a simple binomial form $(a+b)^n$ to apply the Binomial Theorem.

3. Applying the Binomial Theorem

Now we need to find the term independent of $x$ in the expansion of ${\left( {x^{1/3} - x^{-1/2}} \right)^{10}}$. Here, $a = x^{1/3}$, $b = -x^{-1/2}$, and $n=10$.

Step 3.1: Write the general term ($T_{r+1}$) Using the Binomial Theorem formula $T_{r+1} = {}^{n}C_r a^{n-r} b^r$: $T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r$ Explanation: We correctly identify 'a', 'b', and 'n' from our simplified expression. Note that $b$ includes the negative sign.

Step 3.2: Simplify the general term Apply the exponent rules $(x^m)^n = x^{mn}$ and $x^m \cdot x^n = x^{m+n}$: $T_{r+1} = {}^{10}C_r x^{{1 \over 3}(10-r)} (-1)^r x^{-{1 \over 2}r}$ $T_{r+1} = {}^{10}C_r (-1)^r x^{{{10-r} \over 3} - {r \over 2}}$ Explanation: We separated the constant part $(-1)^r$ from the variable part $x$, and combined the exponents of $x$ using the rule for multiplying powers with the same base.

4. Finding the Term Independent of x

For a term to be independent of $x$, its exponent must be zero.

Step 4.1: Set the exponent of x to zero and solve for r Set the exponent of $x$ from the general term to 0: ${{10-r} \over 3} - {r \over 2} = 0$ To eliminate the denominators, multiply the entire equation by the least common multiple of 3 and 2, which is 6: $6 \cdot \left( {{10-r} \over 3} \right) - 6 \cdot \left( {r \over 2} \right) = 6 \cdot 0$ $2(10-r) - 3r = 0$ Distribute and combine like terms: $20 - 2r - 3r = 0$ $20 - 5r = 0$ $5r = 20$ $r = 4$ Explanation: We are looking for the specific term where $x$ effectively disappears, which mathematically means its power is zero. Solving for $r$ tells us which term in the expansion this is.

Step 4.2: Calculate the coefficient of the term Now that we have $r=4$, substitute this value back into the coefficient part of the simplified general term: $\text{Term independent of } x = {}^{10}C_4 (-1)^4$ Calculate ${}^{10}C_4$: ${}^{10}C_4 = {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}$ ${}^{10}C_4 = {{10 \times 9 \times (4 \times 2) \times 7} \over {4 \times 3 \times 2 \times 1}}$ Cancel terms ($4 \times 2$ in the numerator with $4 \times 2$ in the denominator, and $9/3 = 3$): ${}^{10}C_4 = 10 \times 3 \times 7 = 210$ And $(-1)^4 = 1$. Therefore, the term independent of $x$ is: $210 \times 1 = 210$ Explanation: We use the calculated value of $r$ to find the exact binomial coefficient and ensure the sign is correct.

5. Tips and Common Mistakes to Avoid

• Algebraic Simplification First: Always simplify the expression inside the binomial bracket as much as possible before applying the Binomial Theorem. This dramatically reduces complexity.
• Sign Errors: Be very careful with negative signs when identifying the 'b' term in $(a+b)^n$. If it's $(a-b)^n$, then $b$ should be taken as $-b$. In our case, it was $(x^{1/3} - x^{-1/2})^{10}$, so $b = -x^{-1/2}$. Forgetting the negative sign would lead to an incorrect result ($(-1)^r$ factor).
• Exponent Laws: Meticulously apply exponent rules when combining terms with $x$. A common mistake is adding exponents incorrectly or forgetting to multiply exponents when raising a power to another power, i.e., $(x^m)^n \ne x^{m+n}$.
• Checking 'r' value: The value of $r$ must be an integer between $0$ and $n$ (inclusive). If you get a fractional or negative value for $r$, recheck your exponent calculation.

6. Summary and Key Takeaway

The term independent of $x$ in the given expansion is $\mathbf{210}$. The problem highlights a typical strategy: simplify complex algebraic expressions first using identities and exponent rules, then apply general theorems like the Binomial Theorem to find specific terms or coefficients. Pay close attention to signs and exponent manipulation throughout the process.