Expand Using the Binomial Theorem (2a-b)^3

${(2 a - b)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(2 a)}^{3 - k} \cdot {(- b)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply ${(2 a)}^{3}$ by $1$ .

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

Apply the product rule to $2 a$ .

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

Raise $2$ to the power of $3$ .

$8 a^{3} \cdot {(- b)}^{0} + 3 \cdot {(2 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Apply the product rule to $- b$ .

$8 a^{3} \cdot ({(- 1)}^{0} b^{0}) + 3 \cdot {(2 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Rewrite using the commutative property of multiplication.

$8 \cdot {(- 1)}^{0} a^{3} b^{0} + 3 \cdot {(2 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Anything raised to $0$ is $1$ .

$8 \cdot 1 a^{3} b^{0} + 3 \cdot {(2 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Multiply $8$ by $1$ .

$8 a^{3} b^{0} + 3 \cdot {(2 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Anything raised to $0$ is $1$ .

$8 a^{3} \cdot 1 + 3 \cdot {(2 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Multiply $8$ by $1$ .

$8 a^{3} + 3 \cdot {(2 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Apply the product rule to $2 a$ .

$8 a^{3} + 3 \cdot (2^{2} a^{2}) \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

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

$8 a^{3} + 3 \cdot (4 a^{2}) \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Multiply $4$ by $3$ .

$8 a^{3} + 12 a^{2} \cdot {(- b)}^{1} + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Simplify.

$8 a^{3} + 12 a^{2} \cdot (- b) + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Rewrite using the commutative property of multiplication.

$8 a^{3} + 12 \cdot - 1 a^{2} b + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Multiply $12$ by $- 1$ .

$8 a^{3} - 12 a^{2} b + 3 \cdot {(2 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Simplify.

$8 a^{3} - 12 a^{2} b + 3 \cdot (2 a) \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Multiply $2$ by $3$ .

$8 a^{3} - 12 a^{2} b + 6 a \cdot {(- b)}^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Apply the product rule to $- b$ .

$8 a^{3} - 12 a^{2} b + 6 a \cdot ({(- 1)}^{2} b^{2}) + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Rewrite using the commutative property of multiplication.

$8 a^{3} - 12 a^{2} b + 6 \cdot {(- 1)}^{2} a b^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

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

$8 a^{3} - 12 a^{2} b + 6 \cdot 1 a b^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Multiply $6$ by $1$ .

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + 1 \cdot {(2 a)}^{0} \cdot {(- b)}^{3}$

Multiply ${(2 a)}^{0}$ by $1$ .

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + {(2 a)}^{0} \cdot {(- b)}^{3}$

Apply the product rule to $2 a$ .

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + 2^{0} a^{0} \cdot {(- b)}^{3}$

Anything raised to $0$ is $1$ .

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + 1 a^{0} \cdot {(- b)}^{3}$

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

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + a^{0} \cdot {(- b)}^{3}$

Anything raised to $0$ is $1$ .

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + 1 \cdot {(- b)}^{3}$

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

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + {(- b)}^{3}$

Apply the product rule to $- b$ .

$8 a^{3} - 12 a^{2} b + 6 a b^{2} + {(- 1)}^{3} b^{3}$

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

$8 a^{3} - 12 a^{2} b + 6 a b^{2} - b^{3}$

Expand Using the Binomial Theorem (2a-b)^3

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