The coefficient of `x^(-17)` in the expansion of `(x^4-1/x^3)^(15)` is

Explanation

To find the coefficient of \( x^{-17} \) in the expansion of \( (x^4 - \frac{1}{x^3})^{15} \), we will follow these steps:

### Step 1: Apply the Binomial Theorem
According to the Binomial Theorem, we can expand \( (a + b)^n \) as follows:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, let \( a = x^4 \) and \( b = -\frac{1}{x^3} \), and \( n = 15 \). Thus, we have:
\[ (x^4 - \frac{1}{x^3})^{15} = \sum_{k=0}^{15} \binom{15}{k} (x^4)^{15-k} \left(-\frac{1}{x^3}\right)^k \]

### Step 2: Simplify the Terms
Now we simplify the terms in the summation:
\[ = \sum_{k=0}^{15} \binom{15}{k} (-1)^k x^{4(15-k)} \cdot x^{-3k} \]
This can be rewritten as:
\[ = \sum_{k=0}^{15} \binom{15}{k} (-1)^k x^{60 - 7k} \]

### Step 3: Find the Required Power of \( x \)
We need to find the coefficient of \( x^{-17} \). Therefore, we set the exponent equal to -17:
\[ 60 - 7k = -17 \]

### Step 4: Solve for \( k \)
Rearranging the equation gives:
\[ 60 + 17 = 7k \implies 77 = 7k \implies k = \frac{77}{7} = 11 \]

### Step 5: Substitute \( k \) Back into the Coefficient
Now we substitute \( k = 11 \) back into the expression for the coefficient:
\[ \text{Coefficient} = \binom{15}{11} (-1)^{11} \]
Since \( (-1)^{11} = -1 \), we have:
\[ \text{Coefficient} = -\binom{15}{11} \]

### Step 6: Calculate \( \binom{15}{11} \)
Using the property of binomial coefficients, \( \binom{n}{k} = \binom{n}{n-k} \):
\[ \binom{15}{11} = \binom{15}{4} = \frac{15!}{11! \cdot 4!} \]
Calculating this gives:
\[ \binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = \frac{15 \times 14 \times 13 \times 12}{24} \]
Calculating the numerator:
\[ 15 \times 14 = 210, \quad 210 \times 13 = 2730, \quad 2730 \times 12 = 32760 \]
Now divide by 24:
\[ \frac{32760}{24} = 1365 \]

### Step 7: Final Coefficient
Thus, the coefficient of \( x^{-17} \) is:
\[ -\binom{15}{11} = -1365 \]

### Final Answer
The coefficient of \( x^{-17} \) in the expansion of \( (x^4 - \frac{1}{x^3})^{15} \) is \(-1365\).