Expand Using the Binomial Theorem (a-0.5)^3

${(a - 0.5)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(a)}^{3 - k} \cdot {(- 0.5)}^{k}$

Step 2

Expand the summation.

$\frac{3!}{(3 - 0)! 0!} {(a)}^{3 - 0} \cdot {(- 0.5)}^{0} + \frac{3!}{(3 - 1)! 1!} {(a)}^{3 - 1} \cdot {(- 0.5)}^{1} + \frac{3!}{(3 - 2)! 2!} {(a)}^{3 - 2} \cdot {(- 0.5)}^{2} + \frac{3!}{(3 - 3)! 3!} {(a)}^{3 - 3} \cdot {(- 0.5)}^{3}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(a)}^{3} \cdot {(- 0.5)}^{0} + 3 \cdot {(a)}^{2} \cdot {(- 0.5)}^{1} + 3 \cdot {(a)}^{1} \cdot {(- 0.5)}^{2} + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

Step 4

Simplify each term.

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

${(a)}^{3} \cdot {(- 0.5)}^{0} + 3 \cdot {(a)}^{2} \cdot {(- 0.5)}^{1} + 3 \cdot {(a)}^{1} \cdot {(- 0.5)}^{2} + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

Anything raised to $0$ is $1$ .

$a^{3} \cdot 1 + 3 \cdot {(a)}^{2} \cdot {(- 0.5)}^{1} + 3 \cdot {(a)}^{1} \cdot {(- 0.5)}^{2} + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

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

$a^{3} + 3 \cdot {(a)}^{2} \cdot {(- 0.5)}^{1} + 3 \cdot {(a)}^{1} \cdot {(- 0.5)}^{2} + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

Evaluate the exponent.

$a^{3} + 3 a^{2} \cdot - 0.5 + 3 \cdot {(a)}^{1} \cdot {(- 0.5)}^{2} + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

Multiply $- 0.5$ by $3$ .

$a^{3} - 1.5 a^{2} + 3 \cdot {(a)}^{1} \cdot {(- 0.5)}^{2} + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

Simplify.

$a^{3} - 1.5 a^{2} + 3 \cdot a \cdot {(- 0.5)}^{2} + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

Raise $- 0.5$ to the power of $2$ .

$a^{3} - 1.5 a^{2} + 3 a \cdot 0.25 + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

Multiply $0.25$ by $3$ .

$a^{3} - 1.5 a^{2} + 0.75 a + 1 \cdot {(a)}^{0} \cdot {(- 0.5)}^{3}$

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

$a^{3} - 1.5 a^{2} + 0.75 a + {(a)}^{0} \cdot {(- 0.5)}^{3}$

Anything raised to $0$ is $1$ .

$a^{3} - 1.5 a^{2} + 0.75 a + 1 \cdot {(- 0.5)}^{3}$

Multiply ${(- 0.5)}^{3}$ by $1$ .

$a^{3} - 1.5 a^{2} + 0.75 a + {(- 0.5)}^{3}$

Raise $- 0.5$ to the power of $3$ .

$a^{3} - 1.5 a^{2} + 0.75 a - 0.125$

Expand Using the Binomial Theorem (a-0.5)^3

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