Answer
$$(2+3i)^3-(1-i)^3+\frac{5-2i}{2+3i}=\frac{124i-568}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2+3i)^3-(1-i)^3+\frac{5-2i}{2+3i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}8+36i+54i^2+27i^3-(1-3i+3i^2-i^3)+\frac{5-2i}{2+3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}8+36i-54-27i-(1-3i-3+i)+\frac{5-2i}{2+3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}9i-46-(-2i-2)+\frac{5-2i}{2+3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}9i-46-(-2i-2)+\frac{4-19i}{13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}9i-46+2i+2+\frac{4-19i}{13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}11i-44+\frac{4-19i}{13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}\frac{124i-568}{13}\end{aligned} $$ | |
| ① | Find $ \left(2+3i\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 2 $ and $ B = 3i $. $$ \left(2+3i\right)^3 = 2^3+3 \cdot 2^2 \cdot 3i + 3 \cdot 2 \cdot \left( 3i \right)^2+\left( 3i \right)^3 = 8+36i+54i^2+27i^3 $$Find $ \left(1-i\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = 1 $ and $ B = i $. $$ \left(1-i\right)^3 = 1^3-3 \cdot 1^2 \cdot i + 3 \cdot 1 \cdot i^2-i^3 = 1-3i+3i^2-i^3 $$ |
| ② | $$ 54i^2 = 54 \cdot (-1) = -54 $$ |
| ③ | $$ 27i^3 = 27 \cdot \color{blue}{i^2} \cdot i =
27 \cdot ( \color{blue}{-1}) \cdot i =
-27 \cdot \, i $$$$ 3i^2 = 3 \cdot (-1) = -3 $$ |
| ④ | $$ -i^3 = - \color{blue}{i^2} \cdot i =
- ( \color{blue}{-1}) \cdot i =
\, i $$ |
| ⑤ | Combine like terms: $$ \color{blue}{8} + \color{red}{36i} \color{blue}{-54} \color{red}{-27i} = \color{red}{9i} \color{blue}{-46} $$Combine like terms: $$ \color{blue}{1} \color{red}{-3i} \color{blue}{-3} + \color{red}{i} = \color{red}{-2i} \color{blue}{-2} $$ |
| ⑥ | Divide $ \, 5-2i \, $ by $ \, 2+3i \, $ to get $\,\, \dfrac{4-19i}{13} $. ( view steps ) |
| ⑦ | Remove the parentheses by changing the sign of each term within them. $$ - \left( -2i-2 \right) = 2i+2 $$ |
| ⑧ | Combine like terms: $$ \color{blue}{9i} \color{red}{-46} + \color{blue}{2i} + \color{red}{2} = \color{blue}{11i} \color{red}{-44} $$ |
| ⑨ | Add $11i-44$ and $ \dfrac{4-19i}{13} $ to get $ \dfrac{ \color{purple}{ 124i-568 } }{ 13 }$. Step 1: Write $ 11i-44 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To add raitonal expressions, both fractions must have the same denominator. |
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