Find the Product (1-3y)^3

${(1 - 3 y)}^{3}$

Use the Binomial Theorem.

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

Simplify each term.

One to any power is one.

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

One to any power is one.

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

Multiply $3$ by $1$ .

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

Multiply $- 3$ by $3$ .

$1 - 9 y + 3 \cdot 1 {(- 3 y)}^{2} + {(- 3 y)}^{3}$

Multiply $3$ by $1$ .

$1 - 9 y + 3 {(- 3 y)}^{2} + {(- 3 y)}^{3}$

Apply the product rule to $- 3 y$ .

$1 - 9 y + 3 ({(- 3)}^{2} y^{2}) + {(- 3 y)}^{3}$

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

$1 - 9 y + 3 (9 y^{2}) + {(- 3 y)}^{3}$

Multiply $9$ by $3$ .

$1 - 9 y + 27 y^{2} + {(- 3 y)}^{3}$

Apply the product rule to $- 3 y$ .

$1 - 9 y + 27 y^{2} + {(- 3)}^{3} y^{3}$

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

$1 - 9 y + 27 y^{2} - 27 y^{3}$