Question:

<p><strong>Step 1: Write the general term in the expansion of $\left(\dfrac{1}{x^2} - x^3\right)^n$</strong></p><p>Using the binomial theorem: $T_{r+1} = \binom{n}{r}\left(\dfrac{1}{x^2}\right)^{n-r}(-x^3)^r = \binom{n}{r}(-1)^r x^{3r-2(n-r)} = \binom{n}{r}(-1)^r x^{5r-2n}$</p><p><strong>Step 2: Find r for which the power of x equals 5</strong></p><p>For $x^5$: $5r - 2n = 5 \Rightarrow r_1 = \dfrac{2n+5}{5}$</p><p><strong>Step 3: Find r for which the power of x equals 10</strong></p><p>For $x^{10}$: $5r - 2n = 10 \Rightarrow r_2 = \dfrac{2n+10}{5}$</p><p><strong>Step 4: Write the coefficients</strong></p><p>Coefficient of $x^5$: $\binom{n}{r_1}(-1)^{r_1}$</p><p>Coefficient of $x^{10}$: $\binom{n}{r_2}(-1)^{r_2}$</p><p><strong>Step 5: Apply the condition that sum = 0</strong></p><p>$\binom{n}{r_1}(-1)^{r_1} + \binom{n}{r_2}(-1)^{r_2} = 0$</p><p>From Step 2 and 3: $r_2 = r_1 + 1$</p><p>Therefore: $\binom{n}{r_1}(-1)^{r_1} + \binom{n}{r_1+1}(-1)^{r_1+1} = 0$</p><p>$\binom{n}{r_1}(-1)^{r_1} - \binom{n}{r_1+1}(-1)^{r_1} = 0$</p><p>$(-1)^{r_1}\left[\binom{n}{r_1} - \binom{n}{r_1+1}\right] = 0$</p><p>$\binom{n}{r_1} = \binom{n}{r_1+1}$</p><p><strong>Step 6: Solve using the property of binomial coefficients</strong></p><p>$\binom{n}{r_1} = \binom{n}{r_1+1}$ occurs when $r_1 = r_1 + 1$ (impossible) or when $r_1 + (r_1+1) = n$</p><p>$2r_1 + 1 = n$</p><p>Substituting $r_1 = \dfrac{2n+5}{5}$:</p><p>$2 \cdot \dfrac{2n+5}{5} + 1 = n$</p><p>$\dfrac{4n+10}{5} + 1 = n$</p><p>$\dfrac{4n+10+5}{5} = n$</p><p>$4n + 15 = 5n$</p><p>$n = 15$</p><p><strong>∴ Answer:</strong> C</p>