Question
Express the following complex numbers in the standard form a + ib:
$\frac{(1-\text{i})^3}{1-\text{i}^3}$

Answer

$\frac{(1-\text{i})^3}{1-\text{i}^3}=\frac{1^3-\text{i}^3-3\times1\times\text{i}(1+\text{i})}{1-(-\text{i})} \ \begin{bmatrix}\because \ (\text{a}-\text{b})^3=\text{a}^3-\text{b}^3-3\text{ab}(\text{a}+\text{b}) \\ \text{and} \ \text{i}^3=-\text{i} \end{bmatrix}$
$=\frac{1-(-\text{i})-3\text{i}(1-\text{i})}{1+\text{i}}$
$=\frac{1+\text{i}-3\text{i}-3}{1+\text{i}}$
$=\frac{-2-2\text{i}}{1+\text{i}}$
$=\frac{-2(1+\text{i})}{1+\text{i}}$
$=-2$
$=-2+0\text{i}$
$\therefore \ \frac{(1-\text{i})^3}{1-\text{i}^3}=-2+0\text{i}$

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