Answer
$\dfrac{18+6\sqrt{2}}{7}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Rationalize the given expression, $
\dfrac{6}{3-\sqrt{2}}
,$ by multiplying the numerator and the denominator by the conjugate of the denominator.
$\bf{\text{Solution Details:}}$
By reversing the operator between the terms of the denominator, $
3-\sqrt{2}
,$ the conjugate is $
3+\sqrt{2}
.$ Multiplying both the numerator and the denominator by the conjugate results to
\begin{array}{l}\require{cancel}
\dfrac{6}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}}
\\\\=
\dfrac{6(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}
.\end{array}
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}
\dfrac{6(3+\sqrt{2})}{(3)^2-(\sqrt{2})^2}
\\\\=
\dfrac{6(3+\sqrt{2})}{9-2}
\\\\=
\dfrac{6(3+\sqrt{2})}{7}
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{18+6\sqrt{2}}{7}
.\end{array}
