5. (a) Use the binomial theorem to expand $(2-3x)^2$, $|x|<\frac{2}{3}$ - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 6

We have:

$f(x) = \frac{a + bx}{(2-3x)^2}$

To maintain the coefficient of $x$ as zero, we need:

1. For the linear term: $a + b \cdot 0 = 0$, which implies that $b = -\frac{27}{16}$.
2. The condition for $x^2$: $a + b \cdot \frac{27}{16} = \frac{9}{16}$.

Substituting $b$:

$a - \frac{27}{16}=\frac{9}{16}$

Solving for $a$ gives:

$a = \frac{36}{16} = \frac{9}{4}$

Thus,:

• Value of $a = \frac{9}{4}$
• Value of $b = -\frac{27}{16}$

From the expansion:

$(2-3x)^2 = 4 - 6x + 9x^2$

We can deduce that to find the coefficient of $x^3$, we focus on the cubic term in a similar expansion:

The coefficient is obtained from:

$\frac{3a}{16} + \frac{b \cdot 27}{16}$

Calculating:

• Substitute $a = \frac{9}{4}$ and $b = -\frac{27}{16}$:

$\frac{3(9/4)}{16} + \frac{(-27/16) \cdot 27}{16}$

This simplifies to:

• For the term $3a$: $\frac{27}{64}$
• For the term $27b$: $\frac{-729}{256}$

Combining gives:

$\frac{27}{64} - \frac{729}{256} = \frac{S}{T}$ (final simplified fraction)

Thus, the coefficient of $x^3$ is: $\frac{27}{16}$.