Expand Using the Binomial Theorem (x-y)^3

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

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(x)}^{3 - k} \cdot {(- y)}^{k}$

Step 2

Expand the summation.

$\frac{3!}{(3 - 0)! 0!} {(x)}^{3 - 0} \cdot {(- y)}^{0} + \frac{3!}{(3 - 1)! 1!} {(x)}^{3 - 1} \cdot {(- y)}^{1} + \frac{3!}{(3 - 2)! 2!} {(x)}^{3 - 2} \cdot {(- y)}^{2} + \frac{3!}{(3 - 3)! 3!} {(x)}^{3 - 3} \cdot {(- y)}^{3}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{3} \cdot {(- y)}^{0} + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Step 4

Simplify each term.

Multiply ${(x)}^{3}$ by $1$ .

${(x)}^{3} \cdot {(- y)}^{0} + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Apply the product rule to $- y$ .

$x^{3} \cdot ({(- 1)}^{0} y^{0}) + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Rewrite using the commutative property of multiplication.

${(- 1)}^{0} x^{3} y^{0} + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Anything raised to $0$ is $1$ .

$1 x^{3} y^{0} + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Multiply $x^{3}$ by $1$ .

$x^{3} y^{0} + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Anything raised to $0$ is $1$ .

$x^{3} \cdot 1 + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Multiply $x^{3}$ by $1$ .

$x^{3} + 3 \cdot {(x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Simplify.

$x^{3} + 3 x^{2} \cdot (- y) + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Rewrite using the commutative property of multiplication.

$x^{3} + 3 \cdot - 1 x^{2} y + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Multiply $3$ by $- 1$ .

$x^{3} - 3 x^{2} y + 3 \cdot {(x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Simplify.

$x^{3} - 3 x^{2} y + 3 \cdot x \cdot {(- y)}^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Apply the product rule to $- y$ .

$x^{3} - 3 x^{2} y + 3 x \cdot ({(- 1)}^{2} y^{2}) + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Rewrite using the commutative property of multiplication.

$x^{3} - 3 x^{2} y + 3 \cdot {(- 1)}^{2} x y^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

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

$x^{3} - 3 x^{2} y + 3 \cdot 1 x y^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Multiply $3$ by $1$ .

$x^{3} - 3 x^{2} y + 3 x y^{2} + 1 \cdot {(x)}^{0} \cdot {(- y)}^{3}$

Multiply ${(x)}^{0}$ by $1$ .

$x^{3} - 3 x^{2} y + 3 x y^{2} + {(x)}^{0} \cdot {(- y)}^{3}$

Anything raised to $0$ is $1$ .

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

Multiply ${(- y)}^{3}$ 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 $3$ .

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

Expand Using the Binomial Theorem (x-y)^3

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