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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $3 a$ .

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

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

$27 a^{3} \cdot {(4)}^{0} + 3 \cdot {(3 a)}^{2} \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Anything raised to $0$ is $1$ .

$27 a^{3} \cdot 1 + 3 \cdot {(3 a)}^{2} \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Multiply $27$ by $1$ .

$27 a^{3} + 3 \cdot {(3 a)}^{2} \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Apply the product rule to $3 a$ .

$27 a^{3} + 3 \cdot (3^{2} a^{2}) \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Multiply $3$ by $3^{2}$ by adding the exponents.

Move $3^{2}$ .

$27 a^{3} + 3^{2} \cdot 3 \cdot a^{2} \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Multiply $3^{2}$ by $3$ .

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

$27 a^{3} + 3^{2} \cdot 3^{1} \cdot a^{2} \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$27 a^{3} + 3^{2 + 1} \cdot a^{2} \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Add $2$ and $1$ .

$27 a^{3} + 3^{3} \cdot a^{2} \cdot {(4)}^{1} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Simplify $3^{3} \cdot a^{2} \cdot {(4)}^{1}$ .

$27 a^{3} + 3^{3} \cdot a^{2} \cdot 4 + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

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

$27 a^{3} + 27 \cdot a^{2} \cdot 4 + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Multiply $4$ by $27$ .

$27 a^{3} + 108 a^{2} + 3 \cdot {(3 a)}^{1} \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Simplify.

$27 a^{3} + 108 a^{2} + 3 \cdot (3 a) \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Multiply $3$ by $3$ .

$27 a^{3} + 108 a^{2} + 9 a \cdot {(4)}^{2} + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

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

$27 a^{3} + 108 a^{2} + 9 a \cdot 16 + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Multiply $16$ by $9$ .

$27 a^{3} + 108 a^{2} + 144 a + 1 \cdot {(3 a)}^{0} \cdot {(4)}^{3}$

Multiply ${(3 a)}^{0}$ by $1$ .

$27 a^{3} + 108 a^{2} + 144 a + {(3 a)}^{0} \cdot {(4)}^{3}$

Apply the product rule to $3 a$ .

$27 a^{3} + 108 a^{2} + 144 a + 3^{0} a^{0} \cdot {(4)}^{3}$

Anything raised to $0$ is $1$ .

$27 a^{3} + 108 a^{2} + 144 a + 1 a^{0} \cdot {(4)}^{3}$

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

$27 a^{3} + 108 a^{2} + 144 a + a^{0} \cdot {(4)}^{3}$

Anything raised to $0$ is $1$ .

$27 a^{3} + 108 a^{2} + 144 a + 1 \cdot {(4)}^{3}$

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

$27 a^{3} + 108 a^{2} + 144 a + {(4)}^{3}$

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

$27 a^{3} + 108 a^{2} + 144 a + 64$

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

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