Question

The term independent of \(x\) in the expansion of \[ \left(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right)^{15} \] is equal to:

• A. 5105
• B. 5005
• C. 1365
• D. 105

Hint:

Whenever you see fractional exponents like $1/3$ and $2/3$ inside a binomial expression, do not try to expand it directly. It is a reliable sign that the problem contains a hidden cubic or quadratic factoring identity designed to clean up the expression first.

Answer 1

The Correct Option is B

Concept: Before expanding a complicated binomial expression using the Binomial Theorem, look for algebraic factoring identities to simplify the individual fractions first. Specifically, look to apply the sum of cubes and difference of squares formulas: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ $$a^2 - b^2 = (a - b)(a + b)$$ Step 1: Simplify the first fractional term.
Let us rewrite the numerator of the first fraction as a sum of cubes by letting $x = \left(x^{\frac{1}{3}}\right)^3$: $$x + 1 = \left(x^{\frac{1}{3}}\right)^3 + 1^3 = \left(x^{\frac{1}{3}} + 1\right)\left(x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1\right)$$ Substitute this factored expression back into the first fraction: $$\frac{\left(x^{\frac{1}{3}} + 1\right)\left(x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1\right)}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} = x^{\frac{1}{3}} + 1$$

Step 2: Simplify the second fractional term.
Now factor the numerator and denominator of the second fraction using standard square roots:
• Numerator: $x - 1 = \left(x^{\frac{1}{2}}\right)^2 - 1^2 = \left(x^{\frac{1}{2}} - 1\right)\left(x^{\frac{1}{2}} + 1\right)$
• Denominator: $x - x^{\frac{1}{2}} = x^{\frac{1}{2}}\left(x^{\frac{1}{2}} - 1\right)$ Substitute these factored terms back into the second fraction: $$\frac{\left(x^{\frac{1}{2}} - 1\right)\left(x^{\frac{1}{2}} + 1\right)}{x^{\frac{1}{2}}\left(x^{\frac{1}{2}} - 1\right)} = \frac{x^{\frac{1}{2}} + 1}{x^{\frac{1}{2}}} = 1 + \frac{1}{x^{\frac{1}{2}}} = 1 + x^{-\frac{1}{2}}$$

Step 3: Reassemble the simplified binomial expression.
Subtract the two simplified fractions as specified by the brackets: $$\text{Expression} = \left( \left[x^{\frac{1}{3}} + 1\right] - \left[1 + x^{-\frac{1}{2}}\right] \right)^{15} = \left( x^{\frac{1}{3}} - x^{-\frac{1}{2}} \right)^{15}$$

Step 4: Find the term independent of $x$.
The general term $T_{r+1}$ in the binomial expansion of $(A + B)^n$ is given by $T_{r+1} = {}^n C_r \cdot A^{n-r} \cdot B^r$. Substituting our simplified expression parameters: $$T_{r+1} = {}^{15}C_r \cdot \left(x^{\frac{1}{3}}\right)^{15-r} \cdot \left(-x^{-\frac{1}{2}}\right)^r = {}^{15}C_r \cdot (-1)^r \cdot x^{\frac{15-r}{3}} \cdot x^{-\frac{r}{2}}$$ Combine the exponents of $x$: $$\text{Net Exponent} = \frac{15-r}{3} - \frac{r}{2} = 5 - \frac{r}{3} - \frac{r}{2} = 5 - \frac{5r}{6}$$ For the term to be independent of $x$, this net exponent must equal zero: $$5 - \frac{5r}{6} = 0 \quad \Rightarrow \quad 5 = \frac{5r}{6} \quad \Rightarrow \quad r = 6$$

Step 5: Calculate the final numerical coefficient value.
Substitute $r = 6$ back into our general term formula: $$\text{Coefficient} = {}^{15}C_6 \cdot (-1)^6 = {}^{15}C_6 = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 5005$$ This matches option (B) perfectly.