Expand Using the Binomial Theorem (2m+n)^3

${(2 m + n)}^{3}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $2 m$ .

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

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

$8 m^{3} \cdot {(n)}^{0} + 3 \cdot {(2 m)}^{2} \cdot {(n)}^{1} + 3 \cdot {(2 m)}^{1} \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Anything raised to $0$ is $1$ .

$8 m^{3} \cdot 1 + 3 \cdot {(2 m)}^{2} \cdot {(n)}^{1} + 3 \cdot {(2 m)}^{1} \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Multiply $8$ by $1$ .

$8 m^{3} + 3 \cdot {(2 m)}^{2} \cdot {(n)}^{1} + 3 \cdot {(2 m)}^{1} \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Apply the product rule to $2 m$ .

$8 m^{3} + 3 \cdot (2^{2} m^{2}) \cdot {(n)}^{1} + 3 \cdot {(2 m)}^{1} \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

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

$8 m^{3} + 3 \cdot (4 m^{2}) \cdot {(n)}^{1} + 3 \cdot {(2 m)}^{1} \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Multiply $4$ by $3$ .

$8 m^{3} + 12 m^{2} \cdot {(n)}^{1} + 3 \cdot {(2 m)}^{1} \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Simplify.

$8 m^{3} + 12 m^{2} \cdot n + 3 \cdot {(2 m)}^{1} \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Simplify.

$8 m^{3} + 12 m^{2} n + 3 \cdot (2 m) \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Multiply $2$ by $3$ .

$8 m^{3} + 12 m^{2} n + 6 m \cdot {(n)}^{2} + 1 \cdot {(2 m)}^{0} \cdot {(n)}^{3}$

Multiply ${(2 m)}^{0}$ by $1$ .

$8 m^{3} + 12 m^{2} n + 6 m n^{2} + {(2 m)}^{0} \cdot {(n)}^{3}$

Apply the product rule to $2 m$ .

$8 m^{3} + 12 m^{2} n + 6 m n^{2} + 2^{0} m^{0} \cdot {(n)}^{3}$

Anything raised to $0$ is $1$ .

$8 m^{3} + 12 m^{2} n + 6 m n^{2} + 1 m^{0} \cdot {(n)}^{3}$

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

$8 m^{3} + 12 m^{2} n + 6 m n^{2} + m^{0} \cdot {(n)}^{3}$

Anything raised to $0$ is $1$ .

$8 m^{3} + 12 m^{2} n + 6 m n^{2} + 1 \cdot {(n)}^{3}$

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

$8 m^{3} + 12 m^{2} n + 6 m n^{2} + n^{3}$

Expand Using the Binomial Theorem (2m+n)^3

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