Answer
$$\frac{\sqrt{40}^3}{\sqrt{15}^3}=\frac{16\sqrt{6}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{40}^3}{\sqrt{15}^3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{40\sqrt{40}}{15\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{40\sqrt{40}}{15\sqrt{15}}\frac{\sqrt{15}}{\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{400\sqrt{6}}{225} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{ 400 \sqrt{ 6 } : \color{blue}{ 25 } } { 225 : \color{blue}{ 25 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{16\sqrt{6}}{9}\end{aligned} $$ | |
| ① | $$ \sqrt{40}^3 =
\sqrt{40} ^2 \cdot \sqrt{40} =
\lvert 40 \rvert \cdot \sqrt{40} =
40\sqrt{40} $$ |
| ② | $$ \sqrt{15}^3 =
\sqrt{15} ^2 \cdot \sqrt{15} =
\lvert 15 \rvert \cdot \sqrt{15} =
15\sqrt{15} $$ |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{15}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 40 \sqrt{40} } \cdot \sqrt{15} = 400 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ 15 \sqrt{15} } \cdot \sqrt{15} = 225 $$ |
| ⑤ | Divide numerator and denominator by $ \color{blue}{ 25 } $. |
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