The coefficient of `x^(-7)` in the expansion of `(ax - (1)/( bx^(2)))^(8)` is

Options

A. `(-1)^(5)""^(8)C_(5) (a^(3))/( b^(5))`
B. `(-1)^(5)""^(8)C_(5) (a^(5))/( b^(3))`
C. `""^(8)C_(3) (a^(5))/( b^(3))`
D. None of these

Answer

To find the coefficient of \( x^{-7} \) in the expansion of \( (ax - \frac{1}{bx^2})^8 \), we can follow these steps:

### Step 1: Identify the General Term
The general term in the binomial expansion of \( (a x + b y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} (a x)^{n-r} (b y)^r \]
For our expression, we can rewrite it as:
\[ T_{r+1} = \binom{8}{r} (a x)^{8-r} \left(-\frac{1}{b x^2}\right)^r \]

### Step 2: Simplify the General Term
Substituting the values into the general term, we have:
\[ T_{r+1} = \binom{8}{r} (a x)^{8-r} \left(-\frac{1}{b}\right)^r (x^{-2})^r \]
This simplifies to:
\[ T_{r+1} = \binom{8}{r} a^{8-r} (-1)^r b^{-r} x^{(8-r) - 2r} \]
Thus, we can express the power of \( x \) as:
\[ T_{r+1} = \binom{8}{r} a^{8-r} (-1)^r b^{-r} x^{8 - 3r} \]

### Step 3: Set the Power of \( x \) to -7
We want the term where the power of \( x \) is \( -7 \):
\[ 8 - 3r = -7 \]
Solving for \( r \):
\[ 8 + 7 = 3r \implies 15 = 3r \implies r = 5 \]

### Step 4: Substitute \( r \) Back into the General Term
Now, substitute \( r = 5 \) back into the general term:
\[ T_{6} = \binom{8}{5} a^{8-5} (-1)^5 b^{-5} x^{8 - 3 \cdot 5} \]
This becomes:
\[ T_{6} = \binom{8}{5} a^{3} (-1)^5 b^{-5} x^{-7} \]

### Step 5: Find the Coefficient
The coefficient of \( x^{-7} \) is:
\[ \text{Coefficient} = \binom{8}{5} a^{3} (-1)^5 b^{-5} \]
Calculating \( \binom{8}{5} \):
\[ \binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
Thus, the coefficient is:
\[ -56 \cdot \frac{a^3}{b^5} \]

### Final Answer
The coefficient of \( x^{-7} \) in the expansion is:
\[ -\frac{56 a^3}{b^5} \]