We are asked to find the value of $n$ in the expansion of $\left(\frac{1}{x^3} - x^4\right)^n$ given that the sum of the coefficients of $x^7$ and $x^{14}$ is zero.

Binomial Expansion General Term

The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. In this case, $a = \frac{1}{x^3} = x^{-3}$ and $b = -x^4$. So, the general term $T_{r+1}$ is:

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

Simplify the expression:

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

Finding Terms with $x^7$ and $x^{14}$

Let the term containing $x^7$ correspond to $r=r_1$ and the term containing $x^{14}$ correspond to $r=r_2$. For $x^7$, the exponent is 7:

$ 7r_1 - 3n = 7 \quad (1) $ The coefficient is $C_1 = \binom{n}{r_1} (-1)^{r_1}$.

For $x^{14}$, the exponent is 14:

$ 7r_2 - 3n = 14 \quad (2) $ The coefficient is $C_2 = \binom{n}{r_2} (-1)^{r_2}$.

Sum of Coefficients Condition

We are given that the sum of these coefficients is zero:

$ C_1 + C_2 = 0 $ $ \binom{n}{r_1} (-1)^{r_1} + \binom{n}{r_2} (-1)^{r_2} = 0 $

Subtract equation (1) from equation (2):

$ (7r_2 - 3n) - (7r_1 - 3n) = 14 - 7 $ $ 7r_2 - 7r_1 = 7 $ $ r_2 - r_1 = 1 \implies r_2 = r_1 + 1 $

Substitute $r_2 = r_1 + 1$ into the sum of coefficients equation:

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

Divide by $(-1)^{r_1}$ (since it's non-zero):

$ \binom{n}{r_1} - \binom{n}{r_1+1} = 0 $ $ \binom{n}{r_1} = \binom{n}{r_1+1} $

This equality holds if $r_1 = r_1+1$ (which is impossible) or if $r_1 + (r_1+1) = n$. Therefore,

$ n = 2r_1 + 1 $

Solving for $n$

Now substitute $n = 2r_1 + 1$ into equation (1):

$ 7r_1 - 3(2r_1 + 1) = 7 $ $ 7r_1 - 6r_1 - 3 = 7 $ $ r_1 - 3 = 7 $ $ r_1 = 10 $

Finally, find the value of $n$ using $n = 2r_1 + 1$:

$ n = 2(10) + 1 $ $ n = 20 + 1 $ $ n = 21 $

The value of $n$ is 21.