If the sum of the coefficients of $$x^7$$ and $$x^{14}$$ in the expansion of $$\left(\frac{1}{x^3} - x^4\right)^n$$, $$x \neq 0$$, is zero, then the value of n is _______ :

To solve this, we use the general term of the binomial expansion $$(a+b)^n$$, which is $$T_{r+1} = {}^nC_r \cdot a^{n-r} \cdot b^r$$.

For the expansion $$(\frac{1}{x^3} - x^4)^n$$:

$$T_{r+1} = {}^nC_r (x^{-3})^{n-r} (-x^4)^r = {}^nC_r (-1)^r x^{-3n + 3r + 4r} = {}^nC_r (-1)^r x^{7r - 3n}$$

• For $$x^7$$: $$7r_1 - 3n = 7 \implies 7r_1 = 3n + 7 \implies r_1 = \frac{3n+7}{7}$$
• For $$x^{14}$$: $$7r_2 - 3n = 14 \implies 7r_2 = 3n + 14 \implies r_2 = \frac{3n+14}{7}$$

Notice that $$r_2 = r_1 + 1$$.

The coefficients are $${}^nC_{r_1}(-1)^{r_1}$$ and $${}^nC_{r_2}(-1)^{r_2}$$. Given their sum is zero:

$${}^nC_{r_1}(-1)^{r_1} + {}^nC_{r_1+1}(-1)^{r_1+1} = 0$$

Divide by $$(-1)^{r_1}$$:

$${}^nC_{r_1} - {}^nC_{r_1+1} = 0 \implies {}^nC_{r_1} = {}^nC_{r_1+1}$$

In binomial coefficients, $${}^nC_x = {}^nC_y$$ implies either $$x = y$$ (not possible here) or $$x + y = n$$.

$$r_1 + (r_1 + 1) = n$$

$$2r_1 + 1 = n$$

Substitute $$r_1 = \frac{3n+7}{7}$$:

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

$$6n + 14 + 7 = 7n$$

$$21 = n$$

Value of $$n$$ is 21.