Key Concepts and Formulas

This problem requires the application of two main mathematical concepts:

1. Algebraic Factorization Formulas: Specifically, the sum of cubes formula $$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$ and the difference of squares formula $$a^2 - b^2 = (a-b)(a+b)$$. These are crucial for simplifying the complex expression into a manageable binomial form.
2. Binomial Theorem: For an expansion of the form $$(A+B)^n$$, the general term (or the $(r+1)^{th}$ term) is given by: $$T_{r+1} = {^nC_r} A^{n-r} B^r$$ To find the term independent of $x$, we set the power of $x$ in the general term equal to zero.

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Step-by-Step Solution

1. Simplify the Expression Inside the Parentheses

The given expression is $${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$. We will simplify the two fractions separately.

a. Simplify the First Fraction: The first fraction is $${{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}}$$.

• Identify opportunity for sum of cubes: Observe the numerator $x+1$. We can rewrite $x$ as $(x^{1/3})^3$. So, $x+1$ becomes $(x^{1/3})^3 + 1^3$.
• Apply sum of cubes formula: Using $$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$ with $$a = x^{1/3}$$ and $$b = 1$$: $$x+1 = (x^{1/3} + 1)((x^{1/3})^2 - x^{1/3} \cdot 1 + 1^2) = (x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)$$
• Substitute and cancel: Now, substitute this back into the first fraction: $${{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)} \over {{x^{2/3}} - {x^{1/3}} + 1}}$$ The term $(x^{2/3} - x^{1/3} + 1)$ cancels out from the numerator and denominator (assuming $$x^{2/3} - x^{1/3} + 1 \neq 0$$, which is generally true for real $x$). So, the first simplified term is $$x^{1/3} + 1$$.

b. Simplify the Second Fraction: The second fraction is $${{x - 1} \over {x - {x^{1/2}}}}$$.

• Identify opportunity for difference of squares in numerator: The numerator $x-1$ can be written as $(\sqrt{x})^2 - 1^2$.
• Apply difference of squares formula: Using $$a^2 - b^2 = (a-b)(a+b)$$ with $$a = \sqrt{x}$$ and $$b = 1$$: $$x-1 = (\sqrt{x} - 1)(\sqrt{x} + 1)$$
• Factor out common term in denominator: The denominator $$x - x^{1/2}$$ can be factored by taking $$x^{1/2}$$ (or $$\sqrt{x}$$) common: $$x - x^{1/2} = x^{1/2}(x^{1/2} - 1) = \sqrt{x}(\sqrt{x} - 1)$$
• Substitute and cancel: Now, substitute these back into the second fraction: $${{(\sqrt{x} - 1)(\sqrt{x} + 1)} \over {\sqrt{x}(\sqrt{x} - 1)}}$$ The term $(\sqrt{x} - 1)$ cancels out from the numerator and denominator (assuming $$\sqrt{x} - 1 \neq 0$$, i.e., $x \neq 1$). So, the second simplified term is $${{\sqrt{x} + 1} \over {\sqrt{x}}}$$.
• Further simplify the second term: $${{\sqrt{x} + 1} \over {\sqrt{x}}} = {\sqrt{x} \over {\sqrt{x}}} + {1 \over {\sqrt{x}}} = 1 + x^{-1/2}$$.

c. Combine the Simplified Terms: Substitute the simplified first and second terms back into the original expression: $$(x^{1/3} + 1) - (1 + x^{-1/2})$$ $$= x^{1/3} + 1 - 1 - x^{-1/2}$$ $$= x^{1/3} - x^{-1/2}$$ So, the original expression simplifies to $${\left( {x^{1/3} - x^{-1/2}} \right)^{10}}$$.

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2. Apply the Binomial Theorem to Find the Term Independent of $x$

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

a. Write the General Term ($T_{r+1}$): Using the binomial theorem, the general term $$T_{r+1}$$ for $(A+B)^n$ is $$T_{r+1} = {^nC_r} A^{n-r} B^r$$. Here, $$A = x^{1/3}$$, $$B = -x^{-1/2}$$, and $$n = 10$$. Substitute these values into the formula: $$T_{r+1} = {^{10}C_r} (x^{1/3})^{10-r} (-x^{-1/2})^r$$ $$T_{r+1} = {^{10}C_r} x^{{1 \over 3}(10-r)} (-1)^r (x^{-1/2})^r$$ $$T_{r+1} = {^{10}C_r} (-1)^r x^{{10-r} \over 3} x^{-{r \over 2}}$$

• Combine powers of $x$: When multiplying terms with the same base, add their exponents. $$T_{r+1} = {^{10}C_r} (-1)^r x^{({{10-r} \over 3} - {r \over 2})}$$

b. Determine $r$ for the Term Independent of $x$: For the term to be independent of $x$, the exponent of $x$ must be 0. So, we set the exponent equal to zero and solve for $r$: $${{10-r} \over 3} - {r \over 2} = 0$$

• Clear the denominators: Multiply the entire equation by the least common multiple of 3 and 2, which is 6. $$6 \left( {{10-r} \over 3} \right) - 6 \left( {r \over 2} \right) = 6 \cdot 0$$ $$2(10-r) - 3r = 0$$
• Solve for $r$: $$20 - 2r - 3r = 0$$ $$20 - 5r = 0$$ $$5r = 20$$ $$r = 4$$

c. Calculate the Term Independent of $x$: Since $r=4$, the term independent of $x$ is the $(4+1)^{th}$, which is the $5^{th}$ term. Substitute $r=4$ back into the general term $$T_{r+1} = {^{10}C_r} (-1)^r x^{({{10-r} \over 3} - {r \over 2})}$$: $$T_5 = {^{10}C_4} (-1)^4 x^{({{10-4} \over 3} - {4 \over 2})}$$ $$T_5 = {^{10}C_4} (1) x^{({6 \over 3} - 2)}$$ $$T_5 = {^{10}C_4} x^{(2 - 2)}$$ $$T_5 = {^{10}C_4} x^0$$ $$T_5 = {^{10}C_4}$$

d. Calculate the Combination Value: $$^{10}C_4 = {{10!} \over {4!(10-4)!}} = {{10!} \over {4!6!}}$$ $$= {{10 \times 9 \times 8 \times 7 \times 6!} \over {(4 \times 3 \times 2 \times 1) \times 6!}}$$ $$= {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}$$ $$= {{10 \times 9 \times 8 \times 7} \over {24}}$$

• Simplify: $$= 10 \times 3 \times 7$$ (since $9/3 = 3$ and $8/4/2 = 1$) $$= 210$$

Therefore, the term independent of $x$ is 210.

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Tips and Common Mistakes to Avoid

• Algebraic Precision: Be extremely careful with exponents and signs during the initial simplification. A small error here will lead to an incorrect binomial form and an incorrect final answer.
• Recognize Standard Forms: Actively look for patterns that match common algebraic identities like $$a^3 \pm b^3$$ or $$a^2 - b^2$$. These are frequent in competitive exams.
• Powers of $x$: When dealing with terms like $$x^{1/3}$$ and $$x^{-1/2}$$, ensure you correctly apply exponent rules (e.g., $(x^a)^b = x^{ab}$ and $$x^a \cdot x^b = x^{a+b}$$).
• General Term Formula: Memorize and correctly apply the binomial theorem's general term formula. Pay attention to the $$(-1)^r$$ factor if the binomial has a minus sign.
• Calculation of Combinations: Double-check your calculation of $$^nC_r$$ values. Simplify before multiplying large numbers where possible.

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Summary

The problem was solved by first simplifying the complex expression inside the parenthesis using algebraic factorization formulas (sum of cubes and difference of squares) to obtain a simple binomial $${\left( {x^{1/3} - x^{-1/2}} \right)^{10}}$$. Then, the general term of this binomial expansion was found using the binomial theorem. By setting the exponent of $x$ in the general term to zero, we determined the value of $r$ that corresponds to the term independent of $x$. Finally, we calculated the numerical coefficient $$^{10}C_4$$, which yielded 210.