Answer
$5cd^4$
Work Step by Step
Using the $(r+1)$st term of the expansion of $(a+b)^n$, which is given by $\dfrac{n!}{r!(n-r)!}a^{n-r}b^r$, then the $
5
$th term of $
(c-d)^5
$ is
\begin{array}{l}
\dfrac{5!}{4!1!}c^{5-4}d^4
\\\\=
5cd^4
\end{array}