Key Concepts and Formulas

This problem primarily relies on two fundamental mathematical concepts:

1. Algebraic Factorization Identities:
   • Sum of Cubes: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
   • Difference of Squares: $a^2 - b^2 = (a-b)(a+b)$
2. Binomial Theorem: The general term (or $(r+1)^{th}$ term) in the expansion of $(A+B)^n$ is given by: $T_{r+1} = {^nC_r} A^{n-r} B^r$ where ${^nC_r} = {n! \over {r!(n-r)!}}$ is the binomial coefficient.

Step-by-Step Solution

Our goal is to find the coefficient of $x^{-5}$ in the given binomial expansion. The first step is always to simplify the complex expression inside the parentheses to a more manageable form.

1. Simplifying the Base Expression

Let the given expression be $P = \left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}$. We will first simplify the expression inside the large parenthesis. Let's call it $E$.

$E = {{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}$

Simplifying the first fraction: $F_1 = {{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}}$

• Reasoning: We observe that the numerator $x+1$ can be related to the sum of cubes identity. Notice that $x$ can be written as $(x^{1/3})^3$ and $1$ as $(1)^3$.
• Application: Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, with $a = x^{1/3}$ and $b = 1$: $x+1 = (x^{1/3})^3 + (1)^3 = (x^{1/3} + 1)((x^{1/3})^2 - x^{1/3} \cdot 1 + 1^2) = (x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)$
• Cancellation: Now, substitute this back into the fraction: $F_1 = {{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)} \over {x^{2/3} - x^{1/3} + 1}}$ Since $x \ne 0, 1$, the denominator $x^{2/3} - x^{1/3} + 1$ is non-zero, allowing us to cancel the common term: $F_1 = x^{1/3} + 1$

Simplifying the second fraction: $F_2 = {{x - 1} \over {x - {x^{{1 \over 2}}}}}$

• Reasoning (Numerator): The numerator $x-1$ can be expressed using the difference of squares identity. We can write $x$ as $(\sqrt{x})^2 = (x^{1/2})^2$ and $1$ as $(1)^2$.
• Application: Using the identity $a^2 - b^2 = (a-b)(a+b)$, with $a = x^{1/2}$ and $b = 1$: $x-1 = (x^{1/2})^2 - (1)^2 = (x^{1/2} - 1)(x^{1/2} + 1)$
• Reasoning (Denominator): The denominator $x - x^{1/2}$ has a common factor of $x^{1/2}$.
• Factoring: Factor out $x^{1/2}$ from the denominator: $x - x^{1/2} = x^{1/2}(x^{1/2} - 1)$
• Cancellation: Substitute these factored forms back into the fraction: $F_2 = {{(x^{1/2} - 1)(x^{1/2} + 1)} \over {x^{1/2}(x^{1/2} - 1)}}$ Since $x \ne 1$, $x^{1/2} - 1 \ne 0$. We can cancel the common term: $F_2 = {{x^{1/2} + 1} \over {x^{1/2}}}$
• Further Simplification: To simplify further, divide each term in the numerator by the denominator: $F_2 = {{x^{1/2}} \over {x^{1/2}}} + {1 \over {x^{1/2}}} = 1 + {1 \over {x^{1/2}}}$

Combining the simplified fractions: Now, substitute the simplified forms of $F_1$ and $F_2$ back into $E$: $E = (x^{1/3} + 1) - (1 + {1 \over {x^{1/2}}})$

• Reasoning: We remove the parentheses, being careful with the negative sign affecting all terms in the second fraction. $E = x^{1/3} + 1 - 1 - {1 \over {x^{1/2}}}$ $E = x^{1/3} - {1 \over {x^{1/2}}}$
• Reasoning: To prepare for binomial expansion, it's best to express all terms with fractional exponents. We rewrite ${1 \over {x^{1/2}}}$ as $x^{-1/2}$. So, the base expression simplifies to: $E = x^{1/3} - x^{-1/2}$

2. Applying the Binomial Theorem

Now we need to find the coefficient of $x^{-5}$ in the expansion of $(E)^{10} = (x^{1/3} - x^{-1/2})^{10}$. This expression is in the standard binomial form $(A+B)^n$, where:

• $n = 10$ (the power of the entire expression)
• $A = x^{1/3}$ (the first term inside the parenthesis)
• $B = -x^{-1/2}$ (the second term inside the parenthesis, including its sign)

The general term $T_{r+1}$ in the binomial expansion is given by the formula: $T_{r+1} = {^nC_r} A^{n-r} B^r$

• Substitution: Substitute the identified values of $n, A,$ and $B$ into the general term formula: $T_{r+1} = {^{10}C_r} (x^{1/3})^{10-r} (-x^{-1/2})^r$

• Separating Coefficients and Variables: To easily extract the power of $x$, separate the numerical coefficients from the variable parts: $T_{r+1} = {^{10}C_r} (x^{{1 \over 3}(10-r)}) (-1)^r (x^{-{1 \over 2}r})$ $T_{r+1} = {^{10}C_r} (-1)^r x^{{(10-r) \over 3}} x^{{-r \over 2}}$

• Combining Powers of x: Use the exponent rule $x^a x^b = x^{a+b}$ to combine the powers of $x$: $T_{r+1} = {^{10}C_r} (-1)^r x^{{{(10-r)} \over 3} - {r \over 2}}$

• Finding 'r': We are looking for the term where the power of $x$ is $-5$. So, we set the exponent of $x$ equal to $-5$: ${{(10-r)} \over 3} - {r \over 2} = -5$

  • Reasoning: To solve for $r$, we first clear the denominators by multiplying by the least common multiple of 3 and 2, which is 6. $6 \left( {{(10-r)} \over 3} \right) - 6 \left( {r \over 2} \right) = 6(-5)$ $2(10-r) - 3r = -30$
  • Algebraic Simplification: Distribute and combine like terms: $20 - 2r - 3r = -30$ $20 - 5r = -30$
  • Isolate 'r': Subtract 20 from both sides: $-5r = -30 - 20$ $-5r = -50$ Divide by -5: $r = 10$

• Calculating the Coefficient: Now that we have the value of $r = 10$, we can find the coefficient of the term. The term is $T_{r+1} = T_{10+1} = T_{11}$. The coefficient is the part of $T_{r+1}$ that does not involve $x$: Coefficient = ${^{10}C_r} (-1)^r$ Substitute $r = 10$: Coefficient = ${^{10}C_{10}} (-1)^{10}$

  • Reasoning: Recall that ${^nC_n} = 1$ (the number of ways to choose $n$ items from $n$ is 1) and any even power of $-1$ is $1$ (e.g., $(-1)^{10} = 1$). Coefficient = $1 \times 1 = 1$

Thus, the coefficient of $x^{-5}$ in the expansion is 1.

Important Tips and Common Mistakes to Avoid

• Master Algebraic Identities: Proficiency in algebraic factorization identities (like sum/difference of cubes/squares) is critical. Often, complex-looking problems simplify dramatically with their correct application.
• Be Meticulous with Signs: When dealing with negative terms in binomial expansions, such as $(-x^{-1/2})$, ensure that the sign factor $(-1)^r$ is correctly included and evaluated. A common error is to overlook this negative sign.
• Fractional Exponent Rules: A solid understanding of exponent rules, especially with fractional and negative exponents (e.g., $x^{a/b} = \sqrt[b]{x^a}$, $1/x^a = x^{-a}$), is fundamental for accurately combining terms with $x$.
• The Role of 'r': Remember that $r$ in ${^nC_r}$ refers to the index of the term starting from 0, while the term number itself is $(r+1)$. Always verify that your calculated $r$ is a non-negative integer. If $r$ is not an integer, it implies that the desired power of $x$ does not exist in the expansion.

Summary

The problem challenged us to find a specific coefficient in a binomial expansion involving a complex algebraic expression. The solution involved two main stages:

1. Algebraic Simplification: The initial, convoluted expression within the binomial was simplified using the sum of cubes and difference of squares identities, along with careful manipulation of fractional exponents. This transformed the base to a much simpler form: $(x^{1/3} - x^{-1/2})$.
2. Binomial Theorem Application: With the simplified base, we applied the Binomial Theorem. By constructing the general term, equating the exponent of $x$ to the target power of $-5$, we solved for $r$. Finally, substituting the value of $r$ back into the coefficient part of the general term yielded the answer. This problem underscores the necessity of strong foundational algebra skills for success in binomial theorem applications.