## Intermediate Algebra (12th Edition)

$-1+2\sqrt{6}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $[(\sqrt{2}+\sqrt{3})-\sqrt{6}][(\sqrt{2}+\sqrt{3})+\sqrt{6}] ,$ use the special product on multiplying the sum and difference of like terms. $\bf{\text{Solution Details:}}$ Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} (\sqrt{2}+\sqrt{3})^2-(\sqrt{6})^2 \\\\= (\sqrt{2}+\sqrt{3})^2-6 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} [(\sqrt{2})^2+2(\sqrt{2})(\sqrt{3})+(\sqrt{3})^2]-6 \\\\= 2+2(\sqrt{2})(\sqrt{3})+3-6 .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2+2\sqrt{2(3)}+3-6 \\\\= 2+2\sqrt{6}+3-6 .\end{array} Combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (2+3-6)+2\sqrt{6} \\\\= -1+2\sqrt{6} .\end{array}