## Precalculus (6th Edition)

$\color{blue}{\dfrac{1}{2k^2(k-1)}}$
Factor each polynomial to obtain: $=\dfrac{k(k+1)}{4k(2k^2)} \cdot \dfrac{4}{(k+1)(k-1)}$ Cancel the common factors then multiply: $\require{cancel} \\=\dfrac{\cancel{k}\cancel{(k+1)}}{\cancel{4}\cancel{k}(2k^2)} \cdot \dfrac{\cancel{4}}{\cancel{(k+1)}(k-1)} \\=\dfrac{1}{2k^2}\cdot \dfrac{1}{k-1} \\=\color{blue}{\dfrac{1}{2k^2(k-1)}}$