Find the coefficient of
(iii) `(1)/(x^(17) )` in the expansion of `(x^(4) - (1)/(x^3) )^(15)`.

Answer

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

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^4 \), \( b = -\frac{1}{x^3} \), and \( n = 15 \). Therefore, 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, simplifying the general term:
\[ T_{r+1} = \binom{15}{r} (x^{4(15-r)}) \left(-1\right)^r \left(\frac{1}{x^{3r}}\right) \]
This can be rewritten as:
\[ T_{r+1} = \binom{15}{r} (-1)^r x^{60 - 4r - 3r} = \binom{15}{r} (-1)^r x^{60 - 7r} \]

### Step 3: Set the Exponent Equal to -17
We need to find the term where the exponent of \( x \) is \( -17 \):
\[ 60 - 7r = -17 \]
Solving for \( r \):
\[ 60 + 17 = 7r \\ 77 = 7r \\ r = 11 \]

### Step 4: Find the Coefficient
Now, we substitute \( r = 11 \) back into the general term to find the coefficient:
\[ T_{12} = \binom{15}{11} (-1)^{11} x^{60 - 7 \cdot 11} \]
Calculating \( \binom{15}{11} \):
\[ \binom{15}{11} = \binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \]
Thus, the term becomes:
\[ T_{12} = 1365 (-1)^{11} x^{-17} = -1365 x^{-17} \]

### Final Answer
The coefficient of \( \frac{1}{x^{17}} \) in the expansion is:
\[ \text{Coefficient} = -1365 \]