College Algebra, Chapter 9, 9.6, Section 9.6, Problem 42

Determine the term that does not contain $x$ in the expansion of $\displaystyle \left( 8x + \frac{1}{2x} \right)^8$

The term that contains $a^r$ in the expansion of $(a + b)^n$ is

$\displaystyle \left( \begin{array}{c}
n \\
n - r
\end{array} \right) a^r b^{n - r}$

In our case, we have $\displaystyle a = 8x, b = \frac{1}{2x}$ and $n = 8$, so


$
\begin{equation}
\begin{aligned}

=& \left( \begin{array}{c}
8 \\
8 - r
\end{array} \right) (8x)^r \left( \frac{1}{2x} \right)^{8 - r}
\\
\\
=& \left( \begin{array}{c}
8 \\
8 - r
\end{array} \right) (8x)^r \left( \frac{1}{2} x^{-1} \right)^{8 - r}
\\
\\
=& \left( \begin{array}{c}
8 \\
8 - r
\end{array} \right) (8x)^r \left( \frac{1}{2} \right)^{8 - r} (x ^{-1})^{8 - r}
\\
\\
=& \left( \begin{array}{c}
8 \\
8 - r
\end{array} \right) (8x)^r \left( \frac{1}{2} \right)^{8 - r} (x)^{r - 8}
\\
\\
=& \left( \begin{array}{c}
8 \\
8 - r
\end{array} \right) \left( \frac{8^r}{2^{8 - r}} \right) (x)^{2r - 8}

\end{aligned}
\end{equation}
$


Thus, the term that do not contain $x$ is the term in which


$
\begin{equation}
\begin{aligned}

2r - 8 =& 0
\\
r =& 4

\end{aligned}
\end{equation}
$


So, the required term is


$
\begin{equation}
\begin{aligned}

=& \left( \begin{array}{c}
8 \\
8 - 4
\end{array} \right) \left( \frac{8^4}{2^{8 - 4}} \right) (x)^{2 (4) - 4}
\\
\\
=& 17920

\end{aligned}
\end{equation}
$