#### Answer

$8-2\sqrt{12}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To simplify the given radical expression, $
(\sqrt{6}-\sqrt{2})^2
,$ use the special product on squaring binomials and the properties of radicals. Then combine like terms.
$\bf{\text{Solution Details:}}$
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{6})^2-2(\sqrt{6})(\sqrt{2})+(\sqrt{2})^2
\\\\=
6-2(\sqrt{6})(\sqrt{2})+2
.\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}
6-2\sqrt{6(2)}+2
\\\\=
6-2\sqrt{12}+2
.\end{array}
By combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(6+2)-2\sqrt{12}
\\\\=
8-2\sqrt{12}
.\end{array}