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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Anything raised to $0$ is $1$ .

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

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

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

Evaluate the exponent.

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

Multiply $3$ by $3$ .

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

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

Move ${(3)}^{2}$ .

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

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

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

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

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

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

Add $2$ and $1$ .

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

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

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

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

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

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

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

Anything raised to $0$ is $1$ .

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

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

$a^{3} + 9 a^{2} + 27 a + {(3)}^{3}$

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

$a^{3} + 9 a^{2} + 27 a + 27$

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

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