Find the Product (2x-y)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

Apply the product rule to $2 x$ .

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

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

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $- 1$ .

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

Apply the product rule to $2 x$ .

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

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

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

Multiply $4$ by $- 3$ .

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

Multiply $2$ by $3$ .

$8 x^{3} - 12 x^{2} y + 6 x {(- y)}^{2} + {(- y)}^{3}$

Apply the product rule to $- y$ .

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

Rewrite using the commutative property of multiplication.

$8 x^{3} - 12 x^{2} y + 6 \cdot {(- 1)}^{2} x y^{2} + {(- y)}^{3}$

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

$8 x^{3} - 12 x^{2} y + 6 \cdot 1 x y^{2} + {(- y)}^{3}$

Multiply $6$ by $1$ .

$8 x^{3} - 12 x^{2} y + 6 x y^{2} + {(- y)}^{3}$

Apply the product rule to $- y$ .

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

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

$8 x^{3} - 12 x^{2} y + 6 x y^{2} - y^{3}$