Answer
$-1+2\sqrt{6}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
