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

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

Step 1

Remove parentheses.

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

Step 2

Use the Multinomial Theorem.

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

Step 3

Simplify each term.

Multiply $- 2$ by $3$ .

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

Multiply $- 2$ by $6$ .

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

Multiply $- 2$ by $3$ .

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

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

$x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} - 6 x^{2} - 12 x y - 6 y^{2} + 3 x \cdot 4 + 3 y {(- 2)}^{2} + {(- 2)}^{3}$

Multiply $4$ by $3$ .

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

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

$x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} - 6 x^{2} - 12 x y - 6 y^{2} + 12 x + 3 y \cdot 4 + {(- 2)}^{3}$

Multiply $4$ by $3$ .

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

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

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