Simplify (i-2)^3

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

Use the Binomial Theorem.

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

Simplify terms.

Simplify each term.

Factor out $i^{2}$ .

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

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

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

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

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

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

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

Multiply $3$ by $- 1$ .

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

Multiply $- 3$ by $- 2$ .

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

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

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

Multiply $4$ by $3$ .

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

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

$- i + 6 + 12 i - 8$

Simplify by adding terms.

Add $- i$ and $12 i$ .

$6 + 11 i - 8$

Subtract $8$ from $6$ .

$- 2 + 11 i$