Evaluate (1-i)^3

${(1 - i)}^{3}$

Step 1

Use the Binomial Theorem.

$1^{3} + 3 \cdot 1^{2} (- i) + 3 \cdot 1 {(- i)}^{2} + {(- i)}^{3}$

Step 2

Simplify terms.

Simplify each term.

One to any power is one.

$1 + 3 \cdot 1^{2} (- i) + 3 \cdot 1 {(- i)}^{2} + {(- i)}^{3}$

One to any power is one.

$1 + 3 \cdot 1 (- i) + 3 \cdot 1 {(- i)}^{2} + {(- i)}^{3}$

Multiply $3$ by $1$ .

$1 + 3 (- i) + 3 \cdot 1 {(- i)}^{2} + {(- i)}^{3}$

Multiply $- 1$ by $3$ .

$1 - 3 i + 3 \cdot 1 {(- i)}^{2} + {(- i)}^{3}$

Multiply $3$ by $1$ .

$1 - 3 i + 3 {(- i)}^{2} + {(- i)}^{3}$

Apply the product rule to $- i$ .

$1 - 3 i + 3 ({(- 1)}^{2} i^{2}) + {(- i)}^{3}$

Raise $- 1$ to the power of $2$ .

$1 - 3 i + 3 (1 i^{2}) + {(- i)}^{3}$

Multiply $i^{2}$ by $1$ .

$1 - 3 i + 3 i^{2} + {(- i)}^{3}$

Rewrite $i^{2}$ as $- 1$ .

$1 - 3 i + 3 \cdot - 1 + {(- i)}^{3}$

Multiply $3$ by $- 1$ .

$1 - 3 i - 3 + {(- i)}^{3}$

Apply the product rule to $- i$ .

$1 - 3 i - 3 + {(- 1)}^{3} i^{3}$

Raise $- 1$ to the power of $3$ .

$1 - 3 i - 3 - i^{3}$

Factor out $i^{2}$ .

$1 - 3 i - 3 - (i^{2} \cdot i)$

Rewrite $i^{2}$ as $- 1$ .

$1 - 3 i - 3 - (- 1 \cdot i)$

Rewrite $- 1 i$ as $- i$ .

$1 - 3 i - 3 - - i$

Multiply $- 1$ by $- 1$ .

$1 - 3 i - 3 + 1 i$

Multiply $i$ by $1$ .

$1 - 3 i - 3 + i$

Simplify by adding terms.

Subtract $3$ from $1$ .

$- 2 - 3 i + i$

Add $- 3 i$ and $i$ .

$- 2 - 2 i$

Evaluate (1-i)^3

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