The term independent of '\(x\)' in the expansion of \(\left( \frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}} \right)^{10}\), where \(x \neq 0, 1\) is equal to __________.
Show Hint
Always look for factorization in binomial problems with complex bases. Identities like $a^3+b^3$ and $a^2-b^2$ frequently appear in these types of exam questions.
Step 1: Understanding the Concept:
The expression inside the binomial power contains algebraic fractions that can be simplified using factorization identities.
Once simplified, we use the general term formula of the binomial expansion to find the term where the power of \(x\) is zero. Step 2: Key Formula or Approach:
1. Factorization: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\).
2. Factorization: \(a^2 - b^2 = (a-b)(a+b)\).
3. General term of \((A+B)^n\): \(T_{r+1} = \binom{n}{r} A^{n-r} B^r\). Step 3: Detailed Explanation:
Let's simplify the terms inside the bracket:
For the first term:
\[ \frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3} + 1 \]
For the second term:
\[ \frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x})^2 - 1^2}{\sqrt{x}(\sqrt{x}-1)} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2} \]
Now, substituting these back into the original expression:
\[ \left[ (x^{1/3} + 1) - (1 + x^{-1/2}) \right]^{10} = \left( x^{1/3} - x^{-1/2} \right)^{10} \]
The general term \(T_{r+1}\) is given by:
\[ T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r = \binom{10}{r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}} \]
For the term independent of \(x\), the exponent of \(x\) must be zero:
\[ \frac{10-r}{3} - \frac{r}{2} = 0 \implies 20 - 2r - 3r = 0 \implies 5r = 20 \implies r = 4 \]
The coefficient is:
\[ \binom{10}{4} (-1)^4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \] Step 4: Final Answer:
The term independent of \(x\) is 210.
