If the sum of the coefficients of x^7 and x^ 14 in the expansion of...

The general term in the expansion of $\left(\dfrac{1}{x^3} - x^4\right)^n$ is given by:

$T_{r+1} = ^{n}C_{r} \left(x^{-3}\right)^{n-r} \left(-x^4\right)^r = (-1)^r \cdot ^{n}C_{r} \cdot x^{7r - 3n}$

For the coefficient of $x^7$, we set the exponent to $7$:

$7r_1 - 3n = 7 \Rightarrow r_1 = \dfrac{3n + 7}{7}$

For the coefficient of $x^{14}$, we set the exponent to $14$:

$7r_2 - 3n = 14 \Rightarrow r_2 = \dfrac{3n + 14}{7}$

Notice that $r_2 = r_1 + 1$.

The sum of the coefficients of $x^7 and x^{14}$ is zero:

$(-1)^{r_1} \cdot ^{n}C_{r_1} + (-1)^{r_2} \cdot ^{n}C_{r_2} = 0$

$(-1)^{r_1} \cdot ^{n}C_{r_1} + (-1)^{r_1 + 1} \cdot ^{n}C_{r_1 + 1} = 0$

$(-1)^{r_1} \left( ^{n}C_{r_1} - ^{n}C_{r_1 + 1} \right) = 0$

$^{n}C_{r_1} = ^{n}C_{r_1 + 1}$

Since $r_1 \neq r_1 + 1$, we must use the property $^{n}C_{x} = ^{n}C_{y} \Rightarrow x + y = n$:

$r_1 + (r_1 + 1) = n$

$2r_1 + 1 = n$

Substituting $r_1 = \dfrac{3n + 7}{7}$:

$2\left(\dfrac{3n + 7}{7}\right) + 1 = n$

$\dfrac{6n + 14}{7} + 1 = n$

$6n + 14 + 7 = 7n$

$n = 21$

Answer: $21$