$$(2+i)((2+i)^3+\dfrac{4}{3}(2+i)^3i)=\dfrac{44i-117}{3}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2+i)((2+i)^3+\frac{4}{3}(2+i)^3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(2+i)(8+12i+6i^2+i^3+\frac{4}{3}(8+12i+6i^2+i^3)i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(2+i)(8+12i-6-i+\frac{4}{3}(8+12i-6-i)i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}(2+i)(11i+2+\frac{4}{3}(11i+2)i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}(2+i)(11i+2+\frac{44i+8}{3}i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}(2+i)(11i+2+\frac{44i^2+8i}{3}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}(2+i)(11i+2+\frac{-44+8i}{3}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}(2+i)\frac{41i-38}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{41i^2+44i-76}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} } }}}\frac{-41+44i-76}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}\frac{44i-117}{3}\end{aligned} $$ | |
| ① | Find $ \left(2+i\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 2 $ and $ B = i $. $$ \left(2+i\right)^3 = 2^3+3 \cdot 2^2 \cdot i + 3 \cdot 2 \cdot i^2+i^3 = 8+12i+6i^2+i^3 $$Find $ \left(2+i\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 2 $ and $ B = i $. $$ \left(2+i\right)^3 = 2^3+3 \cdot 2^2 \cdot i + 3 \cdot 2 \cdot i^2+i^3 = 8+12i+6i^2+i^3 $$ |
| ② | $$ 6i^2 = 6 \cdot (-1) = -6 $$ |
| ③ | $$ i^3 = \color{blue}{i^2} \cdot i =
( \color{blue}{-1}) \cdot i =
- \, i $$$$ 6i^2 = 6 \cdot (-1) = -6 $$ |
| ④ | $$ i^3 = \color{blue}{i^2} \cdot i =
( \color{blue}{-1}) \cdot i =
- \, i $$ |
| ⑤ | Combine like terms: $$ \color{blue}{8} + \color{red}{12i} \color{blue}{-6} \color{red}{-i} = \color{red}{11i} + \color{blue}{2} $$Combine like terms: $$ \color{blue}{8} + \color{red}{12i} \color{blue}{-6} \color{red}{-i} = \color{red}{11i} + \color{blue}{2} $$ |
| ⑥ | Multiply $ \dfrac{4}{3} $ by $ 11i+2 $ to get $ \dfrac{ 44i+8 }{ 3 } $. Step 1: Write $ 11i+2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{4}{3} \cdot 11i+2 & \xlongequal{\text{Step 1}} \frac{4}{3} \cdot \frac{11i+2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 4 \cdot \left( 11i+2 \right) }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 44i+8 }{ 3 } \end{aligned} $$ |
| ⑦ | Multiply $ \dfrac{44i+8}{3} $ by $ i $ to get $ \dfrac{ 44i^2+8i }{ 3 } $. Step 1: Write $ i $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{44i+8}{3} \cdot i & \xlongequal{\text{Step 1}} \frac{44i+8}{3} \cdot \frac{i}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 44i+8 \right) \cdot i }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 44i^2+8i }{ 3 } \end{aligned} $$ |
| ⑧ | $$ 44i^2 = 44 \cdot (-1) = -44 $$ |
| ⑨ | Add $11i+2$ and $ \dfrac{-44+8i}{3} $ to get $ \dfrac{ \color{purple}{ 41i-38 } }{ 3 }$. Step 1: Write $ 11i+2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To add raitonal expressions, both fractions must have the same denominator. |
| ⑩ | Multiply $2+i$ by $ \dfrac{41i-38}{3} $ to get $ \dfrac{41i^2+44i-76}{3} $. Step 1: Write $ 2+i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2+i \cdot \frac{41i-38}{3} & \xlongequal{\text{Step 1}} \frac{2+i}{\color{red}{1}} \cdot \frac{41i-38}{3} \xlongequal{\text{Step 2}} \frac{ \left( 2+i \right) \cdot \left( 41i-38 \right) }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 82i-76+41i^2-38i }{ 3 } = \frac{41i^2+44i-76}{3} \end{aligned} $$ |
| ⑪ | $$ 41i^2 = 41 \cdot (-1) = -41 $$ |
| ⑫ | Simplify numerator $$ \color{blue}{-41} +44i \color{blue}{-76} = 44i \color{blue}{-117} $$ |