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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $- 2 n$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $- 2$ .

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

Simplify.

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

Apply the product rule to $- 2 n$ .

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

Rewrite using the commutative property of multiplication.

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

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

$m^{3} - 6 m^{2} n + 3 \cdot 4 m n^{2} + 1 \cdot {(m)}^{0} \cdot {(- 2 n)}^{3}$

Multiply $3$ by $4$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

$m^{3} - 6 m^{2} n + 12 m n^{2} + {(- 2 n)}^{3}$

Apply the product rule to $- 2 n$ .

$m^{3} - 6 m^{2} n + 12 m n^{2} + {(- 2)}^{3} n^{3}$

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

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

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

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