Answer
$$\frac{3\sqrt{8}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\frac{12\sqrt{3}-12}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{8}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}& \xlongequal{ }(12\sqrt{3}-12)\cdot\frac{1}{6-2\sqrt{3}+2\sqrt{3}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{12\sqrt{3}-12}{6-2\sqrt{3}+2\sqrt{3}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{3}-12}{4}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \left( 12 \sqrt{3}-12\right) } \cdot 1 = \color{blue}{ 12 \sqrt{3}} \cdot1\color{blue}{-12} \cdot1 = \\ = 12 \sqrt{3}-12 $$$$ \color{blue}{ 1 } \cdot \left( 6- 2 \sqrt{3} + 2 \sqrt{3}-2\right) = \color{blue}{1} \cdot6+\color{blue}{1} \cdot- 2 \sqrt{3}+\color{blue}{1} \cdot 2 \sqrt{3}+\color{blue}{1} \cdot-2 = \\ = 6- 2 \sqrt{3} + 2 \sqrt{3}-2 $$ |
| ② | Simplify numerator and denominator |
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