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

$\sqrt{6}-\sqrt{14}+\sqrt{21}-7$
Using $(a+b)(c+d)=ac+ad+bc+bd$, or the product of 2 binomials, and the properties of radicals, the given expression, $(\sqrt{2}+\sqrt{7})\sqrt{3}-\sqrt{7}) ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt{2}(\sqrt{3})+\sqrt{2}(-\sqrt{7})+\sqrt{7}(\sqrt{3})+\sqrt{7}(-\sqrt{7}) \\\\= \sqrt{2(3)}-\sqrt{2(7)}+\sqrt{7(3)}-\sqrt{7(7)} \\\\= \sqrt{6}-\sqrt{14}+\sqrt{21}-\sqrt{(7)^2} \\\\= \sqrt{6}-\sqrt{14}+\sqrt{21}-7 .\end{array}