Evaluate (x-y)^3

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

Step 1

Use the Binomial Theorem.

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

Step 2

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $- 1$ .

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

Apply the product rule to $- y$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $3$ by $1$ .

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

Apply the product rule to $- y$ .

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

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

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