Precalculus Examples

${(a^{3})}^{- 1} = \frac{1}{216}$

Step 1

Simplify both sides.

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Step 1.1

Multiply the exponents in ${(a^{3})}^{- 1}$ .

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Step 1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a^{3 \cdot - 1} = \frac{1}{216}$

Step 1.1.2

Multiply $3$ by $- 1$ .

$a^{- 3} = \frac{1}{216}$

Step 1.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{a^{3}} = \frac{1}{216}$

Step 2

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$1 \cdot 216 = a^{3} \cdot 1$

Step 3

Solve the equation for $a$ .

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Step 3.1

Rewrite the equation as $a^{3} \cdot 1 = 1 \cdot 216$ .

$a^{3} \cdot 1 = 1 \cdot 216$

Step 3.2

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

$a^{3} = 1 \cdot 216$

Step 3.3

Multiply $216$ by $1$ .

$a^{3} = 216$

Step 3.4

Subtract $216$ from both sides of the equation.

$a^{3} - 216 = 0$

Step 3.5

Factor the left side of the equation.

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Step 3.5.1

Rewrite $216$ as $6^{3}$ .

$a^{3} - 6^{3} = 0$

Step 3.5.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = a$ and $b = 6$ .

$(a - 6) (a^{2} + a \cdot 6 + 6^{2}) = 0$

Step 3.5.3

Simplify.

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Step 3.5.3.1

Move $6$ to the left of $a$ .

$(a - 6) (a^{2} + 6 a + 6^{2}) = 0$

Step 3.5.3.2

Raise $6$ to the power of $2$ .

$(a - 6) (a^{2} + 6 a + 36) = 0$

Step 3.6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$a - 6 = 0$

$a^{2} + 6 a + 36 = 0$

Step 3.7

Set $a - 6$ equal to $0$ and solve for $a$ .

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Step 3.7.1

Set $a - 6$ equal to $0$ .

$a - 6 = 0$

Step 3.7.2

Add $6$ to both sides of the equation.

$a = 6$

Step 3.8

Set $a^{2} + 6 a + 36$ equal to $0$ and solve for $a$ .

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Step 3.8.1

Set $a^{2} + 6 a + 36$ equal to $0$ .

$a^{2} + 6 a + 36 = 0$

Step 3.8.2

Solve $a^{2} + 6 a + 36 = 0$ for $a$ .

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Step 3.8.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.8.2.2

Substitute the values $a = 1$ , $b = 6$ , and $c = 36$ into the quadratic formula and solve for $a$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot 36)}}{2 \cdot 1}$

Step 3.8.2.3

Simplify.

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Step 3.8.2.3.1

Simplify the numerator.

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Step 3.8.2.3.1.1

Raise $6$ to the power of $2$ .

$a = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot 36}}{2 \cdot 1}$

Step 3.8.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 36$ .

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Step 3.8.2.3.1.2.1

Multiply $- 4$ by $1$ .

$a = \frac{- 6 \pm \sqrt{36 - 4 \cdot 36}}{2 \cdot 1}$

Step 3.8.2.3.1.2.2

Multiply $- 4$ by $36$ .

$a = \frac{- 6 \pm \sqrt{36 - 144}}{2 \cdot 1}$

Step 3.8.2.3.1.3

Subtract $144$ from $36$ .

$a = \frac{- 6 \pm \sqrt{- 108}}{2 \cdot 1}$

Step 3.8.2.3.1.4

Rewrite $- 108$ as $- 1 (108)$ .

$a = \frac{- 6 \pm \sqrt{- 1 \cdot 108}}{2 \cdot 1}$

Step 3.8.2.3.1.5

Rewrite $\sqrt{- 1 (108)}$ as $\sqrt{- 1} \cdot \sqrt{108}$ .

$a = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{108}}{2 \cdot 1}$

Step 3.8.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{- 6 \pm i \cdot \sqrt{108}}{2 \cdot 1}$

Step 3.8.2.3.1.7

Rewrite $108$ as $6^{2} \cdot 3$ .

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Step 3.8.2.3.1.7.1

Factor $36$ out of $108$ .

$a = \frac{- 6 \pm i \cdot \sqrt{36 (3)}}{2 \cdot 1}$

Step 3.8.2.3.1.7.2

Rewrite $36$ as $6^{2}$ .

$a = \frac{- 6 \pm i \cdot \sqrt{6^{2} \cdot 3}}{2 \cdot 1}$

Step 3.8.2.3.1.8

Pull terms out from under the radical.

$a = \frac{- 6 \pm i \cdot (6 \sqrt{3})}{2 \cdot 1}$

Step 3.8.2.3.1.9

Move $6$ to the left of $i$ .

$a = \frac{- 6 \pm 6 i \sqrt{3}}{2 \cdot 1}$

Step 3.8.2.3.2

Multiply $2$ by $1$ .

$a = \frac{- 6 \pm 6 i \sqrt{3}}{2}$

Step 3.8.2.3.3

Simplify $\frac{- 6 \pm 6 i \sqrt{3}}{2}$ .

$a = - 3 \pm 3 i \sqrt{3}$

Step 3.8.2.4

The final answer is the combination of both solutions.

$a = - 3 + 3 i \sqrt{3}, - 3 - 3 i \sqrt{3}$

Step 3.9

The final solution is all the values that make $(a - 6) (a^{2} + 6 a + 36) = 0$ true.

$a = 6, - 3 + 3 i \sqrt{3}, - 3 - 3 i \sqrt{3}$