Question:

<p><strong>Step 1: Simplify the first fraction</strong></p><p>Let u = x^(1/3). Then the first fraction becomes:</p><p>$$\frac{x+1}{x^{2/3} - x^{1/3} + 1} = \frac{u^3 + 1}{u^2 - u + 1}$$</p><p>Using the sum of cubes factorization: u³ + 1 = (u + 1)(u² - u + 1)</p><p>$$\frac{(u+1)(u^2 - u + 1)}{u^2 - u + 1} = u + 1 = x^{1/3} + 1$$</p><p><strong>Step 2: Simplify the second fraction</strong></p><p>$$\frac{x-1}{x - x^{1/2}} = \frac{x-1}{x^{1/2}(x^{1/2} - 1)}$$</p><p>Since x - 1 = (x^(1/2) - 1)(x^(1/2) + 1):</p><p>$$\frac{(x^{1/2} - 1)(x^{1/2} + 1)}{x^{1/2}(x^{1/2} - 1)} = \frac{x^{1/2} + 1}{x^{1/2}} = 1 + x^{-1/2}$$</p><p><strong>Step 3: Combine the fractions</strong></p><p>$$\left(x^{1/3} + 1 - 1 - x^{-1/2}\right)^{10} = \left(x^{1/3} - x^{-1/2}\right)^{10}$$</p><p><strong>Step 4: Apply the Binomial Theorem</strong></p><p>The general term is:</p><p>$$T_{r+1} = \binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r = \binom{10}{r}(-1)^r x^{(10-r)/3 - r/2}$$</p><p><strong>Step 5: Find the constant term</strong></p><p>For the constant term, the exponent of x must be zero:</p><p>$$\frac{10-r}{3} - \frac{r}{2} = 0$$</p><p>Multiplying by 6:</p><p>$$2(10-r) - 3r = 0$$</p><p>$$20 - 2r - 3r = 0$$</p><p>$$20 = 5r$$</p><p>$$r = 4$$</p><p><strong>Step 6: Calculate the coefficient</strong></p><p>$$T_5 = \binom{10}{4}(-1)^4 = \binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$$</p><p>$$= \frac{5040}{24} = 210$$</p><p><strong>∴ Answer: 210</strong></p>