Algebra Examples

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify terms.

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the product rule to $2 i$ .

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

Step 2.1.2

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

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

Step 2.1.3

Factor out $i^{2}$ .

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

Step 2.1.4

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

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

Step 2.1.5

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

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

Step 2.1.6

Multiply $- 1$ by $8$ .

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

Step 2.1.7

Apply the product rule to $2 i$ .

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

Step 2.1.8

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

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

Step 2.1.9

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

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

Step 2.1.10

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

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Step 2.1.10.1

Multiply $4$ by $- 1$ .

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

Step 2.1.10.2

Multiply $3$ by $- 4$ .

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

Step 2.1.10.3

Multiply $- 12$ by $- 1$ .

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

Step 2.1.11

Multiply $2$ by $3$ .

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

Step 2.1.12

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

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

Step 2.1.13

Multiply $6$ by $1$ .

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

Step 2.1.14

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

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

Step 2.2

Simplify by adding terms.

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Step 2.2.1

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

$12 - 2 i - 1$

Step 2.2.2

Subtract $1$ from $12$ .

$11 - 2 i$