Answer
$\dfrac{3+3\sqrt{3}}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{\sqrt{27}}{3-\sqrt{3}}
,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to multiply the result.
$\bf{\text{Solution Details:}}$
Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{27}}{3-\sqrt{3}}
\cdot\dfrac{3+\sqrt{3}}{3+\sqrt{3}}
\\\\=
\dfrac{\sqrt{27}(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}
.\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{\sqrt{27}(3+\sqrt{3})}{(3)^2-(\sqrt{3})^2}
\\\\=
\dfrac{\sqrt{27}(3+\sqrt{3})}{9-3}
\\\\=
\dfrac{\sqrt{27}(3+\sqrt{3})}{6}
.\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{\sqrt{27}(3)+\sqrt{27}(\sqrt{3})}{6}
\\\\=
\dfrac{3\sqrt{27}+\sqrt{27}(\sqrt{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}
\dfrac{3\sqrt{27}+\sqrt{27(3)}}{6}
\\\\=
\dfrac{3\sqrt{9\cdot3}+\sqrt{81}}{6}
\\\\=
\dfrac{3\sqrt{(3)^2\cdot3}+\sqrt{(9)^2}}{6}
\\\\=
\dfrac{3(3)\sqrt{3}+9}{6}
\\\\=
\dfrac{9\sqrt{3}+9}{6}
\\\\=
\dfrac{3(3\sqrt{3}+3)}{3\cdot2}
\\\\=
\dfrac{\cancel3(3\sqrt{3}+3)}{\cancel3\cdot2}
\\\\=
\dfrac{3\sqrt{3}+3}{2}
\\\\=
\dfrac{3+3\sqrt{3}}{2}
.\end{array}
