Expressions such as \[ \frac{1}{(1-x)(1+x)} \quad \text{or} \quad \frac{2x+3}{(1-x)(1+2x)} \] are often easier to expand after first splitting them into simpler fractions.

Then each simpler fraction can be expanded using the binomial formula or standard series forms.

1. Decompose the expression into partial fractions.
2. Rewrite each part into a form suitable for binomial expansion, usually \((1+u)^n\).
3. Expand each part to the required power of \(x\).
4. Combine like terms.
5. Read off the coefficient of the required power.

\[ (1+u)^n = 1 + nu + \frac{n(n-1)}{2}u^2 + \frac{n(n-1)(n-2)}{6}u^3 + \cdots \]

Especially useful cases:

\[ (1-u)^{-1} = 1 + u + u^2 + u^3 + \cdots \]

\[ (1+u)^{-1} = 1 - u + u^2 - u^3 + \cdots \]

\[ (1+u)^{-2} = 1 - 2u + 3u^2 - 4u^3 + \cdots \]

Find the coefficient of \(x^3\) in the expansion of

\[ \frac{1}{(1-x)(1+x)}. \]

Step 1: Decompose into partial fractions.

\[ \frac{1}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x} \]

\[ 1=A(1+x)+B(1-x) \]

Let \(x=1\): \(\;1=2A\Rightarrow A=\frac12\)

Let \(x=-1\): \(\;1=2B\Rightarrow B=\frac12\)

So \[ \frac{1}{(1-x)(1+x)}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}. \]

Step 2: Expand each part.

\[ \frac{1}{1-x}=1+x+x^2+x^3+\cdots \]

\[ \frac{1}{1+x}=1-x+x^2-x^3+\cdots \]

Therefore \[ \frac{1}{2(1-x)}+\frac{1}{2(1+x)} = \frac12(1+x+x^2+x^3+\cdots) + \frac12(1-x+x^2-x^3+\cdots) \]

\[ =1+x^2+x^4+\cdots \]

Find the coefficient of \(x^2\) in the expansion of

\[ \frac{3}{(1-x)(1+2x)}. \]

Step 1: Decompose.

\[ \frac{3}{(1-x)(1+2x)}=\frac{A}{1-x}+\frac{B}{1+2x} \]

\[ 3=A(1+2x)+B(1-x) \]

Let \(x=1\): \(\;3=3A\Rightarrow A=1\)

Let \(x=-\frac12\): \(\;3=\frac32 B\Rightarrow B=2\)

So \[ \frac{3}{(1-x)(1+2x)}=\frac{1}{1-x}+\frac{2}{1+2x}. \]

Step 2: Expand.

\[ \frac{1}{1-x}=1+x+x^2+x^3+\cdots \]

\[ \frac{2}{1+2x}=2(1+2x)^{-1} =2(1-2x+4x^2-8x^3+\cdots) \]

\[ =2-4x+8x^2-16x^3+\cdots \]

Add: \[ \frac{3}{(1-x)(1+2x)} = (1+x+x^2+\cdots)+(2-4x+8x^2+\cdots) \]

So the \(x^2\)-term is \[ x^2+8x^2=9x^2. \]

Find the coefficient of \(x^3\) in the expansion of

\[ \frac{1}{(1-x)(1-2x)}. \]

Step 1: Decompose.

\[ \frac{1}{(1-x)(1-2x)}=\frac{A}{1-x}+\frac{B}{1-2x} \]

\[ 1=A(1-2x)+B(1-x) \]

Let \(x=1\): \(\;1=-A\Rightarrow A=-1\)

Let \(x=\frac12\): \(\;1=\frac12 B\Rightarrow B=2\)

Hence \[ \frac{1}{(1-x)(1-2x)}=-\frac{1}{1-x}+\frac{2}{1-2x}. \]

Step 2: Expand.

\[ -\frac{1}{1-x}=-(1+x+x^2+x^3+\cdots) \]

\[ \frac{2}{1-2x}=2(1+2x+4x^2+8x^3+\cdots) \]

\[ =2+4x+8x^2+16x^3+\cdots \]

So the \(x^3\)-term is \[ -x^3+16x^3=15x^3. \]

• After decomposition, each term must be expanded separately.
• Rewrite expressions carefully into the form \((1+u)^n\).
• Be careful with signs, especially in \((1+u)^{-1}\) and \((1-u)^{-1}\).
• If the question asks for a coefficient only, you do not need the full expansion.
• Stop as soon as you have enough terms to identify the required coefficient.