Simplify (2i-1)^3

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

Use the Binomial Theorem.

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

Simplify terms.

Simplify each term.

Apply the product rule to $2 i$ .

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

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

$8 i^{3} + 3 {(2 i)}^{2} \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

Factor out $i^{2}$ .

$8 (i^{2} \cdot i) + 3 {(2 i)}^{2} \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

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

$8 (- 1 \cdot i) + 3 {(2 i)}^{2} \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

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

$8 (- i) + 3 {(2 i)}^{2} \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $- 1$ by $8$ .

$- 8 i + 3 {(2 i)}^{2} \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

Apply the product rule to $2 i$ .

$- 8 i + 3 (2^{2} i^{2}) \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

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

$- 8 i + 3 (4 i^{2}) \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

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

$- 8 i + 3 (4 \cdot - 1) \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $4$ by $- 1$ .

$- 8 i + 3 \cdot - 4 \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $3$ by $- 4$ .

$- 8 i - 12 \cdot - 1 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $- 12$ by $- 1$ .

$- 8 i + 12 + 3 (2 i) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $2$ by $3$ .

$- 8 i + 12 + 6 i {(- 1)}^{2} + {(- 1)}^{3}$

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

$- 8 i + 12 + 6 i \cdot 1 + {(- 1)}^{3}$

Multiply $6$ by $1$ .

$- 8 i + 12 + 6 i + {(- 1)}^{3}$

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

$- 8 i + 12 + 6 i - 1$

Simplify by adding terms.

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

$12 - 2 i - 1$

Subtract $1$ from $12$ .

$11 - 2 i$