Expand using the Binomial Theorem (x-y)^2

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

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(x)}^{2 - k} \cdot {(- y)}^{k}$

Expand the summation.

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

Simplify the exponents for each term of the expansion.

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

Simplify each term.

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Multiply ${(x)}^{2}$ by $1$ .

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

Apply the product rule to $- y$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $2$ by $- 1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

Multiply ${(- y)}^{2}$ by $1$ .

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

Apply the product rule to $- y$ .

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

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

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

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

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

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