Precalculus Examples

${(c + d)}^{3}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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Step 4.1

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

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

Step 4.2

Anything raised to $0$ is $1$ .

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

Step 4.3

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

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

Step 4.4

Simplify.

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

Step 4.5

Simplify.

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

Step 4.6

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

$c^{3} + 3 c^{2} d + 3 c d^{2} + {(c)}^{0} \cdot {(d)}^{3}$

Step 4.7

Anything raised to $0$ is $1$ .

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

Step 4.8

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

$c^{3} + 3 c^{2} d + 3 c d^{2} + d^{3}$