Question
Answer
$$\dfrac{1+3\sqrt{27}}{3\sqrt{-64}}=\dfrac{1+9\sqrt{3}}{24i}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1+3\sqrt{27}}{3\sqrt{-64}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1+9\sqrt{3}}{3\sqrt{64}i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1+9\sqrt{3}}{3\cdot8i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1+9\sqrt{3}}{24i}\end{aligned} $$ | |
| ① | $$ 3 \sqrt{27} =
3 \sqrt{ 3 ^2 \cdot 3 } =
3 \sqrt{ 3 ^2 } \, \sqrt{ 3 } =
3 \cdot 3 \sqrt{ 3 } =
9 \sqrt{ 3 } $$ |
| ② | $$ 3 \sqrt{27} =
3 \sqrt{ 3 ^2 \cdot 3 } =
3 \sqrt{ 3 ^2 } \, \sqrt{ 3 } =
3 \cdot 3 \sqrt{ 3 } =
9 \sqrt{ 3 } $$ |
| ③ | $$ 3 \sqrt{27} =
3 \sqrt{ 3 ^2 \cdot 3 } =
3 \sqrt{ 3 ^2 } \, \sqrt{ 3 } =
3 \cdot 3 \sqrt{ 3 } =
9 \sqrt{ 3 } $$ |
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