In the binomial expansion $${\left( {a + bx} \right)^{ - 3}} = \frac{1}{8} + \frac{9}{8}x + .....,$$      then the value of $$a$$ and $$b$$ are :

Solution :
Given expansion is $${\left( {a + bx} \right)^{ - 3}}$$   which can be written as
$$\eqalign{ & {\left[ {a\left( {1 + \frac{b}{a}x} \right)} \right]^{ - 3}} = {a^{ - 3}}{\left( {1 + \frac{b}{a}x} \right)^{ - 3}} \cr & = {a^{ - 3}}\left( {1 - \frac{{3b}}{a}x + 6{{\left( {\frac{b}{a}x} \right)}^2} - .....} \right) \cr & \left( {{\text{By using}}{{\left( {1 + x} \right)}^{ - 3}} = 1 - 3x + 6{x^2} - .....} \right) \cr} $$
But given that : $${\left( {a + bx} \right)^{ - 3}} = \frac{1}{8} + \frac{9}{8}x + .....$$
$$\eqalign{ & \therefore {a^{ - 3}}\left[ {1 - \frac{{3b}}{a}x + 6\frac{{{b^2}}}{{{a^2}}} \cdot {x^2} - .....} \right] = \frac{1}{8} + \frac{9}{8}x + ..... \cr & \Rightarrow {a^{ - 3}} = \frac{1}{8} = {2^{ - 3}} \cr & \Rightarrow a = 2 \cr & {\text{and }} - 3b{a^{ - 4}} = 9 \cdot {2^{ - 3}} \cr & \Rightarrow b = - 6 \cr} $$