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
- separately expand \((1+x)^\frac{3}{2}\) and \((1-2x)^\frac{3}{4}\) up to \(x^3\) and
- 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
\[\begin{align*}
(1+x)^\frac{3}{2} + (1-2x)^\frac{3}{4} &= \left( 1 + \frac{3}{2} x + \frac{3}{8} x^2 - \frac{3}{48} x^3 \right) + \left( 1 - \frac{3}{2} x - \frac{3}{8} x^2 - \frac{15}{48} x^3 \right)+ \ldots \\
&= 2 - \frac{18}{48} x^3 + \ldots \\
&= 2 - \frac{3}{8} x^3 + \ldots
\end{align*}\]
