## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x^{4}y^{3}\sqrt{y}$
Extracting the factors that are perfect roots of the given index, the given expression, $\sqrt{x^{8}y^{7}} ,$ simplifies to \begin{array}{l}\require{cancel} \sqrt{x^{8}y^{6}\cdot y} \\\\= \sqrt{(x^{4}y^{3})^2\cdot y} \\\\= x^{4}y^{3}\sqrt{y} \end{array} * Note that it is assumed that no radicands were formed by raising negative numbers to even powers.