Find the Product ((x+1)-y)^3

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

Use the Multinomial Theorem.

$x^{3} + 3 x^{2} \cdot 1 + 3 x \cdot 1^{2} + 1^{3} + 3 x^{2} (- y) + 6 x \cdot 1 (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Simplify each term.

Multiply $3$ by $1$ .

$x^{3} + 3 x^{2} + 3 x \cdot 1^{2} + 1^{3} + 3 x^{2} (- y) + 6 x \cdot 1 (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

One to any power is one.

$x^{3} + 3 x^{2} + 3 x \cdot 1 + 1^{3} + 3 x^{2} (- y) + 6 x \cdot 1 (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $3$ by $1$ .

$x^{3} + 3 x^{2} + 3 x + 1^{3} + 3 x^{2} (- y) + 6 x \cdot 1 (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

One to any power is one.

$x^{3} + 3 x^{2} + 3 x + 1 + 3 x^{2} (- y) + 6 x \cdot 1 (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Rewrite using the commutative property of multiplication.

$x^{3} + 3 x^{2} + 3 x + 1 + 3 \cdot - 1 x^{2} y + 6 x \cdot 1 (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $3$ by $- 1$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y + 6 x \cdot 1 (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $6$ by $1$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y + 6 x (- y) + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Rewrite using the commutative property of multiplication.

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y + 6 \cdot - 1 x y + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $6$ by $- 1$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y + 3 \cdot 1^{2} (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

One to any power is one.

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y + 3 \cdot 1 (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $3$ by $1$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y + 3 (- y) + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $- 1$ by $3$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x {(- y)}^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Apply the product rule to $- y$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x ({(- 1)}^{2} y^{2}) + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Rewrite using the commutative property of multiplication.

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 \cdot {(- 1)}^{2} x y^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

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

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 \cdot 1 x y^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $3$ by $1$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x y^{2} + 3 \cdot 1 {(- y)}^{2} + {(- y)}^{3}$

Multiply $3$ by $1$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x y^{2} + 3 {(- y)}^{2} + {(- y)}^{3}$

Apply the product rule to $- y$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x y^{2} + 3 ({(- 1)}^{2} y^{2}) + {(- y)}^{3}$

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

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x y^{2} + 3 (1 y^{2}) + {(- y)}^{3}$

Multiply $y^{2}$ by $1$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x y^{2} + 3 y^{2} + {(- y)}^{3}$

Apply the product rule to $- y$ .

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x y^{2} + 3 y^{2} + {(- 1)}^{3} y^{3}$

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

$x^{3} + 3 x^{2} + 3 x + 1 - 3 x^{2} y - 6 x y - 3 y + 3 x y^{2} + 3 y^{2} - y^{3}$

Find the Product ((x+1)-y)^3

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