Question
Answer
$$\dfrac{(3+i)^3}{3-i}=\dfrac{14+48i}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(3+i)^3}{3-i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{27+27i+9i^2+i^3}{3-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{27+27i-9-i}{3-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{26i+18}{3-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{14+48i}{5}\end{aligned} $$ | |
| ① | Find $ \left(3+i\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 3 $ and $ B = i $. $$ \left(3+i\right)^3 = 3^3+3 \cdot 3^2 \cdot i + 3 \cdot 3 \cdot i^2+i^3 = 27+27i+9i^2+i^3 $$ |
| ② | $$ 9i^2 = 9 \cdot (-1) = -9 $$$$ i^3 = \color{blue}{i^2} \cdot i =
( \color{blue}{-1}) \cdot i =
- \, i $$ |
| ③ | Simplify numerator $$ \color{blue}{27} + \color{red}{27i} \color{blue}{-9} \color{red}{-i} = \color{red}{26i} + \color{blue}{18} $$ |
| ④ | Divide $ \, 18+26i \, $ by $ \, 3-i \, $ to get $\,\, \dfrac{14+48i}{5} $. ( view steps ) |
This page was created using
Complex Expression calculator