Find $ \left(x+iy\right)^3 $ using formula

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

where $ A = x $ and $ B = iy $.

$$ \left(x+iy\right)^3 = x^3+3 \cdot x^2 \cdot iy + 3 \cdot x \cdot \left( iy \right)^2+\left( iy \right)^3 = x^3+3ix^2y+3i^2xy^2+i^3y^3 $$

Subtract $ \dfrac{x^3+3ix^2y+3i^2xy^2+i^3y^3}{3} $ from $ x+iy $ to get $ \dfrac{ \color{purple}{ -i^3y^3-3i^2xy^2-3ix^2y-x^3+3iy+3x } }{ 3 }$.

Step 1: Write $ x+iy $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 3 }$.

$$ \begin{aligned} x+iy- \frac{x^3+3ix^2y+3i^2xy^2+i^3y^3}{3} & \xlongequal{\text{Step 1}} \frac{x+iy}{\color{red}{1}} - \frac{x^3+3ix^2y+3i^2xy^2+i^3y^3}{3} = \\[1ex] &= \frac{ \left( x+iy \right) \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} - \frac{ x^3+3ix^2y+3i^2xy^2+i^3y^3 }{ 3 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x+3iy } }{ 3 } - \frac{ \color{purple}{ x^3+3ix^2y+3i^2xy^2+i^3y^3 } }{ 3 }=\frac{ \color{purple}{ 3x+3iy - \left( x^3+3ix^2y+3i^2xy^2+i^3y^3 \right) } }{ 3 } = \\[1ex] &=\frac{ \color{purple}{ -i^3y^3-3i^2xy^2-3ix^2y-x^3+3iy+3x } }{ 3 } \end{aligned} $$