Answer
\[1-\frac{3}{2}x+\frac{9}{8}{{x}^{2}}+\cdots \]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{\left( 1+\frac{x}{2} \right)}^{-3}} \\
& \text{Using the definition of Taylor series for a function}\left( page\,685 \right) \\
& \sum\limits_{k=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}\text{ } \\
& =f\left( a \right)+f'\left( a \right)\left( x-a \right)+\frac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\cdots \text{ }\left( \mathbf{1} \right) \\
& \text{Where a Taylor series centered at }0\text{ is called a Maclaurin series} \\
& \\
& \text{First}\text{, we calculate some derivatives and their value at }x=0 \\
& f\left( x \right)={{\left( 1+\frac{x}{2} \right)}^{-3}}\to f\left( 0 \right)={{\left( 1+\frac{0}{2} \right)}^{-3}}=1 \\
& f'\left( x \right)=-3{{\left( 1+\frac{x}{2} \right)}^{-3}}\left( \frac{1}{2} \right)\to f'\left( 0 \right)=-3{{\left( 1+\frac{0}{2} \right)}^{-3}}\left( \frac{1}{2} \right)=-\frac{3}{2} \\
& f''\left( x \right)=\frac{9}{2}{{\left( 1+\frac{x}{2} \right)}^{-3}}\left( \frac{1}{2} \right)\to f''\left( 0 \right)=\frac{9}{2}{{\left( 1+\frac{0}{2} \right)}^{-3}}\left( \frac{1}{2} \right)=\frac{9}{4} \\
& \\
& \text{Substituting the previous result into the formula }\left( \mathbf{1} \right). \\
& =1-\frac{3}{2}\left( x-0 \right)+\frac{9/4}{2!}{{\left( x-0 \right)}^{2}}+\cdots \\
& \text{Simplifying} \\
& =1-\frac{3}{2}x+\frac{9}{8}{{x}^{2}}+\cdots \\
\end{align}\]
