Answer
$$\frac{\sqrt{2}^3}{\sqrt{9}^3}=\frac{2\sqrt{2}}{27}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}^3}{\sqrt{9}^3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{2}}{9\sqrt{9}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{2}}{9\sqrt{9}}\frac{\sqrt{9}}{\sqrt{9}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{6\sqrt{2}}{81} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{ 6 \sqrt{ 2 } : \color{blue}{ 3 } } { 81 : \color{blue}{ 3 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{2\sqrt{2}}{27}\end{aligned} $$ | |
| ① | $$ \sqrt{2}^3 =
\sqrt{2} ^2 \cdot \sqrt{2} =
\lvert 2 \rvert \cdot \sqrt{2} =
2\sqrt{2} $$ |
| ② | $$ \sqrt{9}^3 =
\sqrt{9} ^2 \cdot \sqrt{9} =
\lvert 9 \rvert \cdot \sqrt{9} =
9\sqrt{9} $$ |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{9}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{2} } \cdot \sqrt{9} = 6 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ 9 \sqrt{9} } \cdot \sqrt{9} = 81 $$ |
| ⑤ | Divide numerator and denominator by $ \color{blue}{ 3 } $. |
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