Answer
$\dfrac{1}{2k^2(k-1)}$
Work Step by Step
Factoring the expressions and then cancelling the common factors between the numerator and the denominator, the given expression, $
\dfrac{k^2+k}{8k^3}\cdot\dfrac{4}{k^2-1}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{k(k+1)}{8k^3}\cdot\dfrac{4}{(k+1)(k-1)}
\\\\=
\dfrac{\cancel{k}(\cancel{k+1})}{\cancel{4}(2)\cancel{k}\cdot k^2}\cdot\dfrac{\cancel{4}}{(\cancel{k+1})(k-1)}
\\\\=
\dfrac{1}{2k^2(k-1)}
.\end{array}
