Expand Using the Binomial Theorem (10-5b)^3

${(10 - 5 b)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(10)}^{3 - k} \cdot {(- 5 b)}^{k}$

Step 2

Expand the summation.

$\frac{3!}{(3 - 0)! 0!} \cdot {(10)}^{3 - 0} \cdot {(- 5 b)}^{0} + \frac{3!}{(3 - 1)! 1!} \cdot {(10)}^{3 - 1} \cdot {(- 5 b)}^{1} + \frac{3!}{(3 - 2)! 2!} \cdot {(10)}^{3 - 2} \cdot {(- 5 b)}^{2} + \frac{3!}{(3 - 3)! 3!} \cdot {(10)}^{3 - 3} \cdot {(- 5 b)}^{3}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(10)}^{3} \cdot {(- 5 b)}^{0} + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Step 4

Simplify each term.

Multiply ${(10)}^{3}$ by $1$ .

${(10)}^{3} \cdot {(- 5 b)}^{0} + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Raise $10$ to the power of $3$ .

$1000 \cdot {(- 5 b)}^{0} + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Apply the product rule to $- 5 b$ .

$1000 \cdot ({(- 5)}^{0} b^{0}) + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Anything raised to $0$ is $1$ .

$1000 \cdot (1 b^{0}) + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

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

$1000 \cdot b^{0} + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Anything raised to $0$ is $1$ .

$1000 \cdot 1 + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Multiply $1000$ by $1$ .

$1000 + 3 \cdot {(10)}^{2} \cdot {(- 5 b)}^{1} + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Simplify.

$1000 + 3 \cdot {(10)}^{2} \cdot (- 5 b) + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Multiply $- 5$ by $3$ .

$1000 - 15 \cdot {(10)}^{2} b + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Move the decimal point in $- 15$ to the left by $1$ place and increase the power of ${(10)}^{2}$ by $1$ .

$1000 - 1.5 \cdot 10^{3} b + 3 \cdot {(10)}^{1} \cdot {(- 5 b)}^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Apply the product rule to $- 5 b$ .

$1000 - 1.5 \cdot 10^{3} b + 3 \cdot {(10)}^{1} \cdot ({(- 5)}^{2} b^{2}) + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Raise $- 5$ to the power of $2$ .

$1000 - 1.5 \cdot 10^{3} b + 3 \cdot {(10)}^{1} \cdot (25 b^{2}) + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Multiply $25$ by $3$ .

$1000 - 1.5 \cdot 10^{3} b + 75 \cdot {(10)}^{1} b^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Move the decimal point in $75$ to the left by $1$ place and increase the power of ${(10)}^{1}$ by $1$ .

$1000 - 1.5 \cdot 10^{3} b + 7.5 \cdot 10^{2} b^{2} + 1 \cdot {(10)}^{0} \cdot {(- 5 b)}^{3}$

Multiply ${(10)}^{0}$ by $1$ .

$1000 - 1.5 \cdot 10^{3} b + 7.5 \cdot 10^{2} b^{2} + {(10)}^{0} \cdot {(- 5 b)}^{3}$

Anything raised to $0$ is $1$ .

$1000 - 1.5 \cdot 10^{3} b + 7.5 \cdot 10^{2} b^{2} + 1 \cdot {(- 5 b)}^{3}$

Multiply ${(- 5 b)}^{3}$ by $1$ .

$1000 - 1.5 \cdot 10^{3} b + 7.5 \cdot 10^{2} b^{2} + {(- 5 b)}^{3}$

Apply the product rule to $- 5 b$ .

$1000 - 1.5 \cdot 10^{3} b + 7.5 \cdot 10^{2} b^{2} + {(- 5)}^{3} b^{3}$

Raise $- 5$ to the power of $3$ .

$1000 - 1.5 \cdot 10^{3} b + 7.5 \cdot 10^{2} b^{2} - 125 b^{3}$

Expand Using the Binomial Theorem (10-5b)^3

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