Answer
$$\frac{-8\sqrt{3}}{3\sqrt{11}-\sqrt{6}}=\frac{-24\sqrt{33}-24\sqrt{2}}{93}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-8\sqrt{3}}{3\sqrt{11}-\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-8\sqrt{3}}{3\sqrt{11}-\sqrt{6}}\frac{3\sqrt{11}+\sqrt{6}}{3\sqrt{11}+\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-24\sqrt{33}-24\sqrt{2}}{99+3\sqrt{66}-3\sqrt{66}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-24\sqrt{33}-24\sqrt{2}}{93}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{11} + \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ - 8 \sqrt{3} } \cdot \left( 3 \sqrt{11} + \sqrt{6}\right) = \color{blue}{- 8 \sqrt{3}} \cdot 3 \sqrt{11}\color{blue}{- 8 \sqrt{3}} \cdot \sqrt{6} = \\ = - 24 \sqrt{33}- 24 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{11}- \sqrt{6}\right) } \cdot \left( 3 \sqrt{11} + \sqrt{6}\right) = \color{blue}{ 3 \sqrt{11}} \cdot 3 \sqrt{11}+\color{blue}{ 3 \sqrt{11}} \cdot \sqrt{6}\color{blue}{- \sqrt{6}} \cdot 3 \sqrt{11}\color{blue}{- \sqrt{6}} \cdot \sqrt{6} = \\ = 99 + 3 \sqrt{66}- 3 \sqrt{66}-6 $$ |
| ③ | Simplify numerator and denominator |
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