Simplify (-1b-2)^3

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

Step 1

Rewrite $- 1 b$ as $- b$ .

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

Step 2

Use the Binomial Theorem.

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

Step 3

Simplify each term.

Apply the product rule to $- b$ .

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

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

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

Apply the product rule to $- b$ .

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

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

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

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

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

Multiply $- 2$ by $3$ .

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

Multiply $- 1$ by $3$ .

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

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

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

Multiply $4$ by $- 3$ .

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

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

$- b^{3} - 6 b^{2} - 12 b - 8$