Expand Using the Binomial Theorem (1+x)^3

${(1 + x)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(1)}^{3 - k} \cdot {(x)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply $1$ by ${(1)}^{3}$ by adding the exponents.

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

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

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

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

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

Add $1$ and $3$ .

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

Simplify $1^{4} \cdot {(x)}^{0}$ .

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

One to any power is one.

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

One to any power is one.

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

Multiply $3$ by $1$ .

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

Simplify.

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

Evaluate the exponent.

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

Multiply $3$ by $1$ .

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

Multiply $1$ by ${(1)}^{0}$ by adding the exponents.

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

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

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

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

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

Add $1$ and $0$ .

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

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

$1 + 3 x + 3 x^{2} + x^{3}$

Expand Using the Binomial Theorem (1+x)^3

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