Question
Answer
$$\dfrac{(2i^6)^5-{i^5}^2}{i^3}=-\dfrac{31}{-i}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(2i^6)^5-{i^5}^2}{i^3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{32i^{30}-i^{10}}{i^3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-32-(-1)}{i^3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-32+1}{i^3} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{-31}{i^3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-\frac{31}{-i}\end{aligned} $$ | |
| ① | $$ \left( 2i^6 \right)^5 = 2^5 \left( i^6 \right)^5 = 32i^{30} $$ |
| ② | $$ \left( i^5 \right)^2 = 1^2 \left( i^5 \right)^2 = i^{10} $$ |
| ③ | $$ 32i^{30} = 32 \cdot i^{4 \cdot 7 + 2} =
32 \cdot \left( i^4 \right)^{ 7 } \cdot i^2 =
32 \cdot 1^{ 7 } \cdot (-1) =
32 \cdot -1 = -32 $$ |
| ④ | $$ i^{10} = i^{4 \cdot 2 + 2} =
\left( i^4 \right)^{ 2 } \cdot i^2 =
1^{ 2 } \cdot (-1) =
-1 = -1 $$ |
| ⑤ | $ - \, ( \, -1 \, ) = 1 $ |
| ⑥ | $$ i^3 = \color{blue}{i^2} \cdot i =
( \color{blue}{-1}) \cdot i =
- \, i $$ |
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