Find $ \left(4+i\right)^3 $ using formula

$$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$

where $ A = 4 $ and $ B = i $.

$$ \left(4+i\right)^3 = 4^3+3 \cdot 4^2 \cdot i + 3 \cdot 4 \cdot i^2+i^3 = 64+48i+12i^2+i^3 $$

Find $ \left(2-2i\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ 2 } $ and $ B = \color{red}{ 2i }$.

$$ \begin{aligned}\left(2-2i\right)^2 = \color{blue}{2^2} -2 \cdot 2 \cdot 2i + \color{red}{\left( 2i \right)^2} = 4-8i+4i^2\end{aligned} $$

$$ 12i^2 = 12 \cdot (-1) = -12 $$

$$ i^3 = \color{blue}{i^2} \cdot i = ( \color{blue}{-1}) \cdot i = - \, i $$

$$ 4i^2 = 4 \cdot (-1) = -4 $$

Combine like terms:

$$ \color{blue}{64} + \color{red}{48i} \color{blue}{-12} \color{red}{-i} = \color{red}{47i} + \color{blue}{52} $$

Combine like terms:

$$ \, \color{blue}{ \cancel{4}} \,-8i \, \color{blue}{ -\cancel{4}} \, = -8i $$

Divide $ \, 52+47i \, $ by $ \, -8i \, $ to get $\,\, \dfrac{-47+52i}{8} $.
( view steps )

Multiply $i$ by $ \dfrac{-47+52i}{8} $ to get $ \dfrac{52i^2-47i}{8} $.

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.

$$ \begin{aligned} i \cdot \frac{-47+52i}{8} & \xlongequal{\text{Step 1}} \frac{i}{\color{red}{1}} \cdot \frac{-47+52i}{8} \xlongequal{\text{Step 2}} \frac{ i \cdot \left( -47+52i \right) }{ 1 \cdot 8 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -47i+52i^2 }{ 8 } = \frac{52i^2-47i}{8} \end{aligned} $$

$$ 52i^2 = 52 \cdot (-1) = -52 $$