Calculus Examples

$\frac{1}{{(2 i)}^{3}} = {(e^{i x} + e^{- i x})}^{3}$

Step 1

Rewrite the equation as ${(e^{i x} + e^{- i x})}^{3} = \frac{1}{{(2 i)}^{3}}$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{{(2 i)}^{3}}$

Step 2

Simplify $\frac{1}{{(2 i)}^{3}}$ .

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Step 2.1

Simplify the denominator.

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Step 2.1.1

Apply the product rule to $2 i$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{2^{3} i^{3}}$

Step 2.1.2

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

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{8 i^{3}}$

Step 2.1.3

Factor out $i^{2}$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{8 (i^{2} \cdot i)}$

Step 2.1.4

Rewrite $i^{2}$ as $- 1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{8 (- 1 \cdot i)}$

Step 2.1.5

Rewrite $- 1 i$ as $- i$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{8 \cdot - 1 i}$

Step 2.1.6

Combine exponents.

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Step 2.1.6.1

Factor out negative.

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{- (8 i)}$

Step 2.1.6.2

Multiply $8$ by $- 1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{- 8 i}$

Step 2.2

Multiply the numerator and denominator of $\frac{1}{- 8 i}$ by the conjugate of $- 8 i$ to make the denominator real.

${(e^{i x} + e^{- i x})}^{3} = \frac{1}{- 8 i} \cdot \frac{i}{i}$

Step 2.3

Multiply.

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Step 2.3.1

Combine.

${(e^{i x} + e^{- i x})}^{3} = \frac{1 i}{- 8 i i}$

Step 2.3.2

Multiply $i$ by $1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{- 8 i i}$

Step 2.3.3

Simplify the denominator.

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Step 2.3.3.1

Add parentheses.

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{- 8 (i i)}$

Step 2.3.3.2

Raise $i$ to the power of $1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{- 8 (i^{1} i)}$

Step 2.3.3.3

Raise $i$ to the power of $1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{- 8 (i^{1} i^{1})}$

Step 2.3.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{- 8 i^{1 + 1}}$

Step 2.3.3.5

Add $1$ and $1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{- 8 i^{2}}$

Step 2.3.3.6

Rewrite $i^{2}$ as $- 1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{- 8 \cdot - 1}$

Step 2.4

Multiply $- 8$ by $- 1$ .

${(e^{i x} + e^{- i x})}^{3} = \frac{i}{8}$