Question:

<p><strong>Step 1: Simplify the first fraction</strong></p><p>For $\dfrac{x+1}{x^{2/3}-x^{1/3}+1}$, note that $x+1 = (x^{1/3})^3 + 1^3$.</p><p>Using the sum of cubes formula: $a^3+b^3 = (a+b)(a^2-ab+b^2)$</p><p>$$x+1 = (x^{1/3}+1)(x^{2/3}-x^{1/3}+1)$$</p><p>Therefore: $\dfrac{x+1}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$</p><p><strong>Step 2: Simplify the second fraction</strong></p><p>For $\dfrac{x-1}{x-x^{1/2}}$, factor the denominator:</p><p>$$x-x^{1/2} = x^{1/2}(x^{1/2}-1)$$</p><p>Note that $x-1 = (x^{1/2})^2 - 1 = (x^{1/2}-1)(x^{1/2}+1)$</p><p>Therefore: $\dfrac{x-1}{x-x^{1/2}} = \dfrac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)} = \dfrac{x^{1/2}+1}{x^{1/2}} = 1 + x^{-1/2}$</p><p><strong>Step 3: Combine the simplified expressions</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>Using $(a+b)^{10} = \sum_{r=0}^{10} \binom{10}{r} a^{10-r}b^r$, with $a = x^{1/3}$ and $b = -x^{-1/2}$:</p><p>$$\left(x^{1/3} - x^{-1/2}\right)^{10} = \sum_{r=0}^{10} \binom{10}{r} (x^{1/3})^{10-r}(-x^{-1/2})^r$$</p><p>$$= \sum_{r=0}^{10} \binom{10}{r}(-1)^r x^{(10-r)/3 - r/2}$$</p><p><strong>Step 5: Find the power of x that equals -5</strong></p><p>We need: $\dfrac{10-r}{3} - \dfrac{r}{2} = -5$</p><p>Multiply by 6: $2(10-r) - 3r = -30$</p><p>$$20 - 2r - 3r = -30$$</p><p>$$20 - 5r = -30$$</p><p>$$5r = 50$$</p><p>$$r = 10$$</p><p><strong>Step 6: Calculate the coefficient</strong></p><p>The coefficient of $x^{-5}$ is: $\binom{10}{10}(-1)^{10} = 1 \cdot 1 = 1$</p><p><strong>∴ Answer: A</strong></p>