Expand Using the Binomial Theorem (a-b)^4

${(a - b)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(a)}^{4 - k} \cdot {(- b)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} {(a)}^{4 - 0} \cdot {(- b)}^{0} + \frac{4!}{(4 - 1)! 1!} {(a)}^{4 - 1} \cdot {(- b)}^{1} + \frac{4!}{(4 - 2)! 2!} {(a)}^{4 - 2} \cdot {(- b)}^{2} + \frac{4!}{(4 - 3)! 3!} {(a)}^{4 - 3} \cdot {(- b)}^{3} + \frac{4!}{(4 - 4)! 4!} {(a)}^{4 - 4} \cdot {(- b)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(a)}^{4} \cdot {(- b)}^{0} + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Step 4

Simplify each term.

Multiply ${(a)}^{4}$ by $1$ .

${(a)}^{4} \cdot {(- b)}^{0} + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Apply the product rule to $- b$ .

$a^{4} \cdot ({(- 1)}^{0} b^{0}) + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Rewrite using the commutative property of multiplication.

${(- 1)}^{0} a^{4} b^{0} + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Anything raised to $0$ is $1$ .

$1 a^{4} b^{0} + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Multiply $a^{4}$ by $1$ .

$a^{4} b^{0} + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Anything raised to $0$ is $1$ .

$a^{4} \cdot 1 + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Multiply $a^{4}$ by $1$ .

$a^{4} + 4 \cdot {(a)}^{3} \cdot {(- b)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Simplify.

$a^{4} + 4 a^{3} \cdot (- b) + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Rewrite using the commutative property of multiplication.

$a^{4} + 4 \cdot - 1 a^{3} b + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Multiply $4$ by $- 1$ .

$a^{4} - 4 a^{3} b + 6 \cdot {(a)}^{2} \cdot {(- b)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Apply the product rule to $- b$ .

$a^{4} - 4 a^{3} b + 6 a^{2} \cdot ({(- 1)}^{2} b^{2}) + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Rewrite using the commutative property of multiplication.

$a^{4} - 4 a^{3} b + 6 \cdot {(- 1)}^{2} a^{2} b^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

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

$a^{4} - 4 a^{3} b + 6 \cdot 1 a^{2} b^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Multiply $6$ by $1$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} + 4 \cdot {(a)}^{1} \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Simplify.

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} + 4 \cdot a \cdot {(- b)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Apply the product rule to $- b$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} + 4 a \cdot ({(- 1)}^{3} b^{3}) + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Rewrite using the commutative property of multiplication.

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} + 4 \cdot {(- 1)}^{3} a b^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

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

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} + 4 \cdot - 1 a b^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

Multiply $4$ by $- 1$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + 1 \cdot {(a)}^{0} \cdot {(- b)}^{4}$

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

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + {(a)}^{0} \cdot {(- b)}^{4}$

Anything raised to $0$ is $1$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + 1 \cdot {(- b)}^{4}$

Multiply ${(- b)}^{4}$ by $1$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + {(- b)}^{4}$

Apply the product rule to $- b$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + {(- 1)}^{4} b^{4}$

Raise $- 1$ to the power of $4$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + 1 b^{4}$

Multiply $b^{4}$ by $1$ .

$a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + b^{4}$

Expand Using the Binomial Theorem (a-b)^4

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