Find the Product (2x+3y)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

Apply the product rule to $2 x$ .

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

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

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $3$ .

$8 x^{3} + 9 {(2 x)}^{2} y + 3 (2 x) {(3 y)}^{2} + {(3 y)}^{3}$

Apply the product rule to $2 x$ .

$8 x^{3} + 9 (2^{2} x^{2}) y + 3 (2 x) {(3 y)}^{2} + {(3 y)}^{3}$

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

$8 x^{3} + 9 (4 x^{2}) y + 3 (2 x) {(3 y)}^{2} + {(3 y)}^{3}$

Multiply $4$ by $9$ .

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

Multiply $2$ by $3$ .

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

Apply the product rule to $3 y$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $6$ by $9$ .

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

Apply the product rule to $3 y$ .

$8 x^{3} + 36 x^{2} y + 54 x y^{2} + 3^{3} y^{3}$

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

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