Find the Product (2x-3)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

Apply the product rule to $2 x$ .

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

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

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

Apply the product rule to $2 x$ .

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

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

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

Multiply $4$ by $3$ .

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

Multiply $- 3$ by $12$ .

$8 x^{3} - 36 x^{2} + 3 (2 x) {(- 3)}^{2} + {(- 3)}^{3}$

Multiply $2$ by $3$ .

$8 x^{3} - 36 x^{2} + 6 x {(- 3)}^{2} + {(- 3)}^{3}$

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

$8 x^{3} - 36 x^{2} + 6 x \cdot 9 + {(- 3)}^{3}$

Multiply $9$ by $6$ .

$8 x^{3} - 36 x^{2} + 54 x + {(- 3)}^{3}$

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

$8 x^{3} - 36 x^{2} + 54 x - 27$