Solution

Use the binomial theorem to expand \[\begin{equation*} (1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} \end{equation*}\]

as far as the term in \(x^3\).

The binomial theorem says that, if \(|x| < 1\), then \[\begin{equation*} (1+x)^r = 1 + rx + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \dotsb \end{equation*}\] and so to expand \[\begin{equation*} (1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} \end{equation*}\]

as far as the term in \(x^3\), we can

  1. separately expand \((1+x)^\frac{3}{2}\) and \((1-2x)^\frac{3}{4}\) up to \(x^3\) and
  2. collect the terms.
Thus \[\begin{align*} (1+x)^\frac{3}{2} &= 1 + \frac{3}{2}x + \frac{\frac{3}{2} \times \frac{1}{2}}{2!} x^2 + \frac{\frac{3}{2} \times \frac{1}{2} \times \left(- \frac{1}{2} \right)}{3!} x^3 + \ldots\\ &= 1 + \frac{3}{2} x + \frac{3}{8} x^2 - \frac{3}{48} x^3 + \ldots, \end{align*}\] and \[\begin{align*} (1-2x)^\frac{3}{4} &= 1 + \frac{3}{4} (-2x) + \frac{\frac{3}{4} \times \left( -\frac{1}{4} \right)}{2!} (-2x)^2 + \frac{\frac{3}{4} \times \left( -\frac{1}{4} \right) \times \left( -\frac{5}{4} \right)}{3!} (-2x)^3 + \ldots\\ &= 1 - \frac{3}{2} x - \frac{3}{8} x^2 - \frac{15}{48} x^3 + \ldots, \end{align*}\] and therefore \[\begin{align*} (1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} &= \left( 1 + \frac{3}{2} x + \frac{3}{8} x^2 - \frac{3}{48} x^3 \right) + \left( 1 - \frac{3}{2} x - \frac{3}{8} x^2 - \frac{15}{48} x^3 \right)+ \ldots \\ &= 2 - \frac{18}{48} x^3 + \ldots \\ &= 2 - \frac{3}{8} x^3 + \ldots \end{align*}\]