Step 1: Understanding the Concept:
Before expanding using the Binomial Theorem, we must simplify the expression inside the parentheses. The terms look like algebraic identities involving cubes and squares.
Step 2: Key Formula or Approach:
1. Use \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \).
2. Use \( a^2 - b^2 = (a-b)(a+b) \).
Step 3: Detailed Explanation:
Simplify the first term: \( \frac{x+1}{x^{2/3} - x^{1/3} + 1} \).
Let \( t = x^{1/3} \). Then \( x+1 = t^3 + 1 = (t+1)(t^2 - t + 1) \).
The fraction is \( \frac{(x^{1/3}+1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1 \).
Simplify the second term: \( \frac{x-1}{x - x^{1/2}} \).
Let \( u = x^{1/2} \). Then \( x-1 = u^2 - 1 = (u-1)(u+1) \).
And \( x - x^{1/2} = u^2 - u = u(u-1) \).
The fraction is \( \frac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)} = \frac{x^{1/2}+1}{x^{1/2}} = 1 + x^{-1/2} \).
The whole expression becomes:
\[ ( (x^{1/3} + 1) - (1 + x^{-1/2}) )^{15} = (x^{1/3} - x^{-1/2})^{15} \]
The general term \( T_{r+1} = \binom{15}{r} (x^{1/3})^{15-r} (-x^{-1/2})^r \).
\( T_{r+1} = \binom{15}{r} (-1)^r x^{\frac{15-r}{3} - \frac{r}{2}} \).
For the term independent of \( x \), the power of \( x \) must be zero:
\[ \frac{15-r}{3} - \frac{r}{2} = 0 \implies 2(15-r) - 3r = 0 \implies 30 - 2r - 3r = 0 \implies 5r = 30 \implies r = 6 \]
The term is \( \binom{15}{6} (-1)^6 = \binom{15}{6} \).
\[ \binom{15}{6} = \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 5005 \]
Step 4: Final Answer:
The term independent of \( x \) is 5005.