Find the Product (x-3y)^3

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

Use the Binomial Theorem.

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

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $- 3$ .

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

Apply the product rule to $- 3 y$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $3$ by $9$ .

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

Apply the product rule to $- 3 y$ .

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

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

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