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

$3i\sqrt{2}-8i$
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ and $i=\sqrt{-1},$ the given expression is equivalent to \begin{array}{l}\require{cancel} \sqrt{-18}-\sqrt{-64} \\\\= \sqrt{-1}\cdot\sqrt{18}-\sqrt{-1}\cdot\sqrt{64} \\\\= i\sqrt{18}-i\sqrt{64} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} i\sqrt{18}-i\sqrt{64} \\\\= i\sqrt{9\cdot2}-i\sqrt{64} \\\\= i\sqrt{(3)^2\cdot2}-i\sqrt{(8)^2} \\\\= i\cdot3\sqrt{2}-i\cdot8 \\\\= 3i\sqrt{2}-8i \\\\= 3i\sqrt{2}-8i .\end{array}