Expand Using the Binomial Theorem (c+2d)^3

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

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(c)}^{3 - k} \cdot {(2 d)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $2 d$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $2$ .

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

Simplify.

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

Apply the product rule to $2 d$ .

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

Rewrite using the commutative property of multiplication.

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

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

$c^{3} + 6 c^{2} d + 3 \cdot 4 c d^{2} + 1 \cdot {(c)}^{0} \cdot {(2 d)}^{3}$

Multiply $3$ by $4$ .

$c^{3} + 6 c^{2} d + 12 c d^{2} + 1 \cdot {(c)}^{0} \cdot {(2 d)}^{3}$

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

$c^{3} + 6 c^{2} d + 12 c d^{2} + {(c)}^{0} \cdot {(2 d)}^{3}$

Anything raised to $0$ is $1$ .

$c^{3} + 6 c^{2} d + 12 c d^{2} + 1 \cdot {(2 d)}^{3}$

Multiply ${(2 d)}^{3}$ by $1$ .

$c^{3} + 6 c^{2} d + 12 c d^{2} + {(2 d)}^{3}$

Apply the product rule to $2 d$ .

$c^{3} + 6 c^{2} d + 12 c d^{2} + 2^{3} d^{3}$

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

$c^{3} + 6 c^{2} d + 12 c d^{2} + 8 d^{3}$

Expand Using the Binomial Theorem (c+2d)^3

Related Questions

Support LCMGCF.com by disabling your adblocker.