Expand Using the Binomial Theorem (d-4b)^3

${(d - 4 b)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(d)}^{3 - k} \cdot {(- 4 b)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $- 4 b$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $- 4$ .

$d^{3} - 12 d^{2} b + 3 \cdot {(d)}^{1} \cdot {(- 4 b)}^{2} + 1 \cdot {(d)}^{0} \cdot {(- 4 b)}^{3}$

Simplify.

$d^{3} - 12 d^{2} b + 3 \cdot d \cdot {(- 4 b)}^{2} + 1 \cdot {(d)}^{0} \cdot {(- 4 b)}^{3}$

Apply the product rule to $- 4 b$ .

$d^{3} - 12 d^{2} b + 3 d \cdot ({(- 4)}^{2} b^{2}) + 1 \cdot {(d)}^{0} \cdot {(- 4 b)}^{3}$

Rewrite using the commutative property of multiplication.

$d^{3} - 12 d^{2} b + 3 \cdot {(- 4)}^{2} d b^{2} + 1 \cdot {(d)}^{0} \cdot {(- 4 b)}^{3}$

Raise $- 4$ to the power of $2$ .

$d^{3} - 12 d^{2} b + 3 \cdot 16 d b^{2} + 1 \cdot {(d)}^{0} \cdot {(- 4 b)}^{3}$

Multiply $3$ by $16$ .

$d^{3} - 12 d^{2} b + 48 d b^{2} + 1 \cdot {(d)}^{0} \cdot {(- 4 b)}^{3}$

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

$d^{3} - 12 d^{2} b + 48 d b^{2} + {(d)}^{0} \cdot {(- 4 b)}^{3}$

Anything raised to $0$ is $1$ .

$d^{3} - 12 d^{2} b + 48 d b^{2} + 1 \cdot {(- 4 b)}^{3}$

Multiply ${(- 4 b)}^{3}$ by $1$ .

$d^{3} - 12 d^{2} b + 48 d b^{2} + {(- 4 b)}^{3}$

Apply the product rule to $- 4 b$ .

$d^{3} - 12 d^{2} b + 48 d b^{2} + {(- 4)}^{3} b^{3}$

Raise $- 4$ to the power of $3$ .

$d^{3} - 12 d^{2} b + 48 d b^{2} - 64 b^{3}$

Expand Using the Binomial Theorem (d-4b)^3

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