The sum of the coefficients of `x^(32)` and `x^(-17)` in `(x^4- 1/(x^3))^15` is

Explanation

To solve the problem of finding the sum of the coefficients of \(x^{32}\) and \(x^{-17}\) in the expansion of \((x^4 - \frac{1}{x^3})^{15}\), we will follow these steps:

### Step 1: Identify the General Term
The general term in the binomial expansion of \((p + q)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]

In our case, \(p = x^4\), \(q = -\frac{1}{x^3}\), and \(n = 15\). Thus, the general term becomes:

\[ T_{r+1} = \binom{15}{r} (x^4)^{15-r} \left(-\frac{1}{x^3}\right)^r \]

### Step 2: Simplify the General Term
Now, we simplify \(T_{r+1}\):

\[ T_{r+1} = \binom{15}{r} x^{4(15-r)} \left(-1\right)^r \frac{1}{x^{3r}} = \binom{15}{r} (-1)^r x^{60 - 4r - 3r} = \binom{15}{r} (-1)^r x^{60 - 7r} \]

### Step 3: Find the Coefficient of \(x^{32}\)
To find the coefficient of \(x^{32}\), we set the exponent equal to 32:

\[ 60 - 7r = 32 \]

Solving for \(r\):

\[ 7r = 60 - 32 \implies 7r = 28 \implies r = 4 \]

Now, we find the coefficient corresponding to \(r = 4\):

\[ \text{Coefficient of } x^{32} = \binom{15}{4} (-1)^4 = \binom{15}{4} \]

### Step 4: Find the Coefficient of \(x^{-17}\)
Next, we find the coefficient of \(x^{-17}\):

\[ 60 - 7r = -17 \]

Solving for \(r\):

\[ 7r = 60 + 17 \implies 7r = 77 \implies r = 11 \]

Now, we find the coefficient corresponding to \(r = 11\):

\[ \text{Coefficient of } x^{-17} = \binom{15}{11} (-1)^{11} = -\binom{15}{11} \]

### Step 5: Sum the Coefficients
Now we sum the coefficients of \(x^{32}\) and \(x^{-17}\):

\[ \text{Sum} = \binom{15}{4} - \binom{15}{11} \]

Using the property of binomial coefficients \(\binom{n}{r} = \binom{n}{n-r}\):

\[ \binom{15}{11} = \binom{15}{4} \]

Thus, we have:

\[ \text{Sum} = \binom{15}{4} - \binom{15}{4} = 0 \]

### Final Answer
The sum of the coefficients of \(x^{32}\) and \(x^{-17}\) is:

\[ \boxed{0} \]