Expand Using the Binomial Theorem (6+3z)^3

${(6 + 3 z)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(6)}^{3 - k} \cdot {(3 z)}^{k}$

Step 2

Expand the summation.

$\frac{3!}{(3 - 0)! 0!} \cdot {(6)}^{3 - 0} \cdot {(3 z)}^{0} + \frac{3!}{(3 - 1)! 1!} \cdot {(6)}^{3 - 1} \cdot {(3 z)}^{1} + \frac{3!}{(3 - 2)! 2!} \cdot {(6)}^{3 - 2} \cdot {(3 z)}^{2} + \frac{3!}{(3 - 3)! 3!} \cdot {(6)}^{3 - 3} \cdot {(3 z)}^{3}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(6)}^{3} \cdot {(3 z)}^{0} + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Step 4

Simplify each term.

Multiply ${(6)}^{3}$ by $1$ .

${(6)}^{3} \cdot {(3 z)}^{0} + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Raise $6$ to the power of $3$ .

$216 \cdot {(3 z)}^{0} + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Apply the product rule to $3 z$ .

$216 \cdot (3^{0} z^{0}) + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Anything raised to $0$ is $1$ .

$216 \cdot (1 z^{0}) + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

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

$216 \cdot z^{0} + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Anything raised to $0$ is $1$ .

$216 \cdot 1 + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Multiply $216$ by $1$ .

$216 + 3 \cdot {(6)}^{2} \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Raise $6$ to the power of $2$ .

$216 + 3 \cdot 36 \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Multiply $3$ by $36$ .

$216 + 108 \cdot {(3 z)}^{1} + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Simplify.

$216 + 108 \cdot (3 z) + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Multiply $3$ by $108$ .

$216 + 324 z + 3 \cdot {(6)}^{1} \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Evaluate the exponent.

$216 + 324 z + 3 \cdot 6 \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Multiply $3$ by $6$ .

$216 + 324 z + 18 \cdot {(3 z)}^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Apply the product rule to $3 z$ .

$216 + 324 z + 18 \cdot (3^{2} z^{2}) + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

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

$216 + 324 z + 18 \cdot (9 z^{2}) + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Multiply $9$ by $18$ .

$216 + 324 z + 162 z^{2} + 1 \cdot {(6)}^{0} \cdot {(3 z)}^{3}$

Multiply ${(6)}^{0}$ by $1$ .

$216 + 324 z + 162 z^{2} + {(6)}^{0} \cdot {(3 z)}^{3}$

Anything raised to $0$ is $1$ .

$216 + 324 z + 162 z^{2} + 1 \cdot {(3 z)}^{3}$

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

$216 + 324 z + 162 z^{2} + {(3 z)}^{3}$

Apply the product rule to $3 z$ .

$216 + 324 z + 162 z^{2} + 3^{3} z^{3}$

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

$216 + 324 z + 162 z^{2} + 27 z^{3}$

Expand Using the Binomial Theorem (6+3z)^3

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