Expand Using the Binomial Theorem (h-k)^4

${(h - k)}^{4}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $- k$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $4$ by $- 1$ .

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

Apply the product rule to $- k$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $6$ by $1$ .

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

Simplify.

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

Apply the product rule to $- k$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $4$ by $- 1$ .

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

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

$h^{4} - 4 h^{3} k + 6 h^{2} k^{2} - 4 h k^{3} + {(h)}^{0} \cdot {(- k)}^{4}$

Anything raised to $0$ is $1$ .

$h^{4} - 4 h^{3} k + 6 h^{2} k^{2} - 4 h k^{3} + 1 \cdot {(- k)}^{4}$

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

$h^{4} - 4 h^{3} k + 6 h^{2} k^{2} - 4 h k^{3} + {(- k)}^{4}$

Apply the product rule to $- k$ .

$h^{4} - 4 h^{3} k + 6 h^{2} k^{2} - 4 h k^{3} + {(- 1)}^{4} k^{4}$

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

$h^{4} - 4 h^{3} k + 6 h^{2} k^{2} - 4 h k^{3} + 1 k^{4}$

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

$h^{4} - 4 h^{3} k + 6 h^{2} k^{2} - 4 h k^{3} + k^{4}$

Expand Using the Binomial Theorem (h-k)^4

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