In the binomial expansion `(a+bx)^(-3) = ( 1)/( 8 ) + (9)/( 8)x +"…….." ,` then the value of a and b are `:`

Options

A. a=2,b =3
B. a=2,b = -6
C. a=3,b =2
D. a= -3, b = 2

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) in the binomial expansion of \( (a + bx)^{-3} \) given that it equals \( \frac{1}{8} + \frac{9}{8}x + \ldots \).

### Step-by-Step Solution:

1. **Identify the Binomial Expansion Form**:
The binomial expansion for \( (a + bx)^{-3} \) can be expressed using the binomial theorem:
\[ (a + bx)^{-3} = a^{-3} \left(1 + \frac{bx}{a}\right)^{-3} \]
Using the binomial expansion formula, we have:
\[ (1 + u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \ldots \]
where \( u = \frac{bx}{a} \) and \( n = 3 \).

2. **Rewrite the Expansion**:
Thus, we can write:
\[ (a + bx)^{-3} = a^{-3} \left(1 - 3\frac{bx}{a} + \frac{3 \cdot 4}{2!}\left(\frac{bx}{a}\right)^2 - \ldots\right) \]
This simplifies to:
\[ = \frac{1}{a^3} - \frac{3b}{a^4}x + \ldots \]

3. **Compare with Given Expansion**:
We are given:
\[ \frac{1}{8} + \frac{9}{8}x + \ldots \]
From the comparison, we can equate the constant terms and the coefficients of \( x \):
- From the constant term:
\[ \frac{1}{a^3} = \frac{1}{8} \implies a^3 = 8 \implies a = 2 \]
- From the coefficient of \( x \):
\[ -\frac{3b}{a^4} = \frac{9}{8} \]
Substituting \( a = 2 \):
\[ -\frac{3b}{2^4} = \frac{9}{8} \implies -\frac{3b}{16} = \frac{9}{8} \]

4. **Solve for \( b \)**:
To isolate \( b \), multiply both sides by -16:
\[ 3b = -16 \cdot \frac{9}{8} \implies 3b = -18 \implies b = -6 \]

5. **Final Values**:
We have found:
\[ a = 2, \quad b = -6 \]

### Conclusion:
The values of \( a \) and \( b \) are:
\[ \boxed{(2, -6)} \]