Expand Using the Binomial Theorem (8-m)^3

${(8 - m)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(8)}^{3 - k} \cdot {(- m)}^{k}$

Step 2

Expand the summation.

$\frac{3!}{(3 - 0)! 0!} \cdot {(8)}^{3 - 0} \cdot {(- m)}^{0} + \frac{3!}{(3 - 1)! 1!} \cdot {(8)}^{3 - 1} \cdot {(- m)}^{1} + \frac{3!}{(3 - 2)! 2!} \cdot {(8)}^{3 - 2} \cdot {(- m)}^{2} + \frac{3!}{(3 - 3)! 3!} \cdot {(8)}^{3 - 3} \cdot {(- m)}^{3}$

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply ${(8)}^{3}$ by $1$ .

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

Raise $8$ to the power of $3$ .

$512 \cdot {(- m)}^{0} + 3 \cdot {(8)}^{2} \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Apply the product rule to $- m$ .

$512 \cdot ({(- 1)}^{0} m^{0}) + 3 \cdot {(8)}^{2} \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Anything raised to $0$ is $1$ .

$512 \cdot (1 m^{0}) + 3 \cdot {(8)}^{2} \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

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

$512 \cdot m^{0} + 3 \cdot {(8)}^{2} \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Anything raised to $0$ is $1$ .

$512 \cdot 1 + 3 \cdot {(8)}^{2} \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Multiply $512$ by $1$ .

$512 + 3 \cdot {(8)}^{2} \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Raise $8$ to the power of $2$ .

$512 + 3 \cdot 64 \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Multiply $3$ by $64$ .

$512 + 192 \cdot {(- m)}^{1} + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Simplify.

$512 + 192 \cdot (- m) + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Multiply $- 1$ by $192$ .

$512 - 192 m + 3 \cdot {(8)}^{1} \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Evaluate the exponent.

$512 - 192 m + 3 \cdot 8 \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Multiply $3$ by $8$ .

$512 - 192 m + 24 \cdot {(- m)}^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Apply the product rule to $- m$ .

$512 - 192 m + 24 \cdot ({(- 1)}^{2} m^{2}) + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Raise $- 1$ to the power of $2$ .

$512 - 192 m + 24 \cdot (1 m^{2}) + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Multiply $m^{2}$ by $1$ .

$512 - 192 m + 24 \cdot m^{2} + 1 \cdot {(8)}^{0} \cdot {(- m)}^{3}$

Multiply ${(8)}^{0}$ by $1$ .

$512 - 192 m + 24 m^{2} + {(8)}^{0} \cdot {(- m)}^{3}$

Anything raised to $0$ is $1$ .

$512 - 192 m + 24 m^{2} + 1 \cdot {(- m)}^{3}$

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

$512 - 192 m + 24 m^{2} + {(- m)}^{3}$

Apply the product rule to $- m$ .

$512 - 192 m + 24 m^{2} + {(- 1)}^{3} m^{3}$

Raise $- 1$ to the power of $3$ .

$512 - 192 m + 24 m^{2} - m^{3}$

Expand Using the Binomial Theorem (8-m)^3

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