Question
Answer
$$0^3\sqrt{-192}=0\sqrt{3}i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}0^3\sqrt{-192}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}0\sqrt{-192} \xlongequal{ } \\[1 em] & \xlongequal{ }0\sqrt{192}i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}0\cdot \sqrt{ 64 \cdot 3 } \, i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}0\cdot \sqrt{ 64 } \cdot \sqrt{ 3 } \, i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}0\cdot8 \sqrt{ 3 } \, i \xlongequal{ } \\[1 em] & \xlongequal{ }0\sqrt{3}i\end{aligned} $$ | |
| ① | $ 0 ^ 3 = 0 $ |
| ② | Factor out the largest perfect square of 192. ( in this example we factored out $ 64 $ ) |
| ③ | Rewrite $ \sqrt{ 64 \cdot 3 } $ as the product of two radicals. |
| ④ | The square root of $ 64 $ is $ 8 $. |
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