Expand Using the Binomial Theorem (4a+b)^3

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

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.

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

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

Apply the product rule to $4 a$ .

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

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

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

Anything raised to $0$ is $1$ .

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

Multiply $64$ by $1$ .

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

Apply the product rule to $4 a$ .

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

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

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

Multiply $16$ by $3$ .

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

Simplify.

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

Simplify.

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

Multiply $4$ by $3$ .

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

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

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

Apply the product rule to $4 a$ .

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Expand Using the Binomial Theorem (4a+b)^3

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