## Intermediate Algebra (12th Edition)

$\dfrac{3+3\sqrt{3}}{2}$
$\bf{\text{Solution Outline:}}$ 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}