Precalculus Examples

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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Step 4.1

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

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

Step 4.2

Apply the product rule to $4 a$ .

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

Step 4.3

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

$64 a^{3} \cdot {(- b)}^{0} + 3 \cdot {(4 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.4

Apply the product rule to $- b$ .

$64 a^{3} \cdot ({(- 1)}^{0} b^{0}) + 3 \cdot {(4 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.5

Rewrite using the commutative property of multiplication.

$64 \cdot {(- 1)}^{0} a^{3} b^{0} + 3 \cdot {(4 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.6

Anything raised to $0$ is $1$ .

$64 \cdot 1 a^{3} b^{0} + 3 \cdot {(4 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.7

Multiply $64$ by $1$ .

$64 a^{3} b^{0} + 3 \cdot {(4 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.8

Anything raised to $0$ is $1$ .

$64 a^{3} \cdot 1 + 3 \cdot {(4 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.9

Multiply $64$ by $1$ .

$64 a^{3} + 3 \cdot {(4 a)}^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.10

Apply the product rule to $4 a$ .

$64 a^{3} + 3 \cdot (4^{2} a^{2}) \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.11

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

$64 a^{3} + 3 \cdot (16 a^{2}) \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.12

Multiply $16$ by $3$ .

$64 a^{3} + 48 a^{2} \cdot {(- b)}^{1} + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.13

Simplify.

$64 a^{3} + 48 a^{2} \cdot (- b) + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.14

Rewrite using the commutative property of multiplication.

$64 a^{3} + 48 \cdot - 1 a^{2} b + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.15

Multiply $48$ by $- 1$ .

$64 a^{3} - 48 a^{2} b + 3 \cdot {(4 a)}^{1} \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.16

Simplify.

$64 a^{3} - 48 a^{2} b + 3 \cdot (4 a) \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.17

Multiply $4$ by $3$ .

$64 a^{3} - 48 a^{2} b + 12 a \cdot {(- b)}^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.18

Apply the product rule to $- b$ .

$64 a^{3} - 48 a^{2} b + 12 a \cdot ({(- 1)}^{2} b^{2}) + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.19

Rewrite using the commutative property of multiplication.

$64 a^{3} - 48 a^{2} b + 12 \cdot {(- 1)}^{2} a b^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.20

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

$64 a^{3} - 48 a^{2} b + 12 \cdot 1 a b^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.21

Multiply $12$ by $1$ .

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + 1 \cdot {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.22

Multiply ${(4 a)}^{0}$ by $1$ .

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + {(4 a)}^{0} \cdot {(- b)}^{3}$

Step 4.23

Apply the product rule to $4 a$ .

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + 4^{0} a^{0} \cdot {(- b)}^{3}$

Step 4.24

Anything raised to $0$ is $1$ .

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + 1 a^{0} \cdot {(- b)}^{3}$

Step 4.25

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

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + a^{0} \cdot {(- b)}^{3}$

Step 4.26

Anything raised to $0$ is $1$ .

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + 1 \cdot {(- b)}^{3}$

Step 4.27

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

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + {(- b)}^{3}$

Step 4.28

Apply the product rule to $- b$ .

$64 a^{3} - 48 a^{2} b + 12 a b^{2} + {(- 1)}^{3} b^{3}$

Step 4.29

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

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