Expand Using the Binomial Theorem (p+4q)^3

${(p + 4 q)}^{3}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(p)}^{3} \cdot {(4 q)}^{0} + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Step 4

Simplify each term.

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

${(p)}^{3} \cdot {(4 q)}^{0} + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Apply the product rule to $4 q$ .

$p^{3} \cdot (4^{0} q^{0}) + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Rewrite using the commutative property of multiplication.

$4^{0} p^{3} q^{0} + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Anything raised to $0$ is $1$ .

$1 p^{3} q^{0} + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

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

$p^{3} q^{0} + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Anything raised to $0$ is $1$ .

$p^{3} \cdot 1 + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

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

$p^{3} + 3 \cdot {(p)}^{2} \cdot {(4 q)}^{1} + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Simplify.

$p^{3} + 3 p^{2} \cdot (4 q) + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Rewrite using the commutative property of multiplication.

$p^{3} + 3 \cdot 4 p^{2} q + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Multiply $3$ by $4$ .

$p^{3} + 12 p^{2} q + 3 \cdot {(p)}^{1} \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Simplify.

$p^{3} + 12 p^{2} q + 3 \cdot p \cdot {(4 q)}^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Apply the product rule to $4 q$ .

$p^{3} + 12 p^{2} q + 3 p \cdot (4^{2} q^{2}) + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Rewrite using the commutative property of multiplication.

$p^{3} + 12 p^{2} q + 3 \cdot 4^{2} p q^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

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

$p^{3} + 12 p^{2} q + 3 \cdot 16 p q^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

Multiply $3$ by $16$ .

$p^{3} + 12 p^{2} q + 48 p q^{2} + 1 \cdot {(p)}^{0} \cdot {(4 q)}^{3}$

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

$p^{3} + 12 p^{2} q + 48 p q^{2} + {(p)}^{0} \cdot {(4 q)}^{3}$

Anything raised to $0$ is $1$ .

$p^{3} + 12 p^{2} q + 48 p q^{2} + 1 \cdot {(4 q)}^{3}$

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

$p^{3} + 12 p^{2} q + 48 p q^{2} + {(4 q)}^{3}$

Apply the product rule to $4 q$ .

$p^{3} + 12 p^{2} q + 48 p q^{2} + 4^{3} q^{3}$

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

$p^{3} + 12 p^{2} q + 48 p q^{2} + 64 q^{3}$

Expand Using the Binomial Theorem (p+4q)^3

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