Simplify (1-2i)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify terms.

Simplify each term.

One to any power is one.

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

One to any power is one.

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

Multiply $3$ by $1$ .

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

Multiply $- 2$ by $3$ .

$1 - 6 i + 3 \cdot 1 {(- 2 i)}^{2} + {(- 2 i)}^{3}$

Multiply $3$ by $1$ .

$1 - 6 i + 3 {(- 2 i)}^{2} + {(- 2 i)}^{3}$

Apply the product rule to $- 2 i$ .

$1 - 6 i + 3 ({(- 2)}^{2} i^{2}) + {(- 2 i)}^{3}$

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

$1 - 6 i + 3 (4 i^{2}) + {(- 2 i)}^{3}$

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

$1 - 6 i + 3 (4 \cdot - 1) + {(- 2 i)}^{3}$

Multiply $3 (4 \cdot - 1)$ .

Multiply $4$ by $- 1$ .

$1 - 6 i + 3 \cdot - 4 + {(- 2 i)}^{3}$

Multiply $3$ by $- 4$ .

$1 - 6 i - 12 + {(- 2 i)}^{3}$

Apply the product rule to $- 2 i$ .

$1 - 6 i - 12 + {(- 2)}^{3} i^{3}$

Raise $- 2$ to the power of $3$ .

$1 - 6 i - 12 - 8 i^{3}$

Factor out $i^{2}$ .

$1 - 6 i - 12 - 8 (i^{2} \cdot i)$

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

$1 - 6 i - 12 - 8 (- 1 \cdot i)$

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

$1 - 6 i - 12 - 8 (- i)$

Multiply $- 1$ by $- 8$ .

$1 - 6 i - 12 + 8 i$

Simplify by adding terms.

Subtract $12$ from $1$ .

$- 11 - 6 i + 8 i$

Add $- 6 i$ and $8 i$ .

$- 11 + 2 i$