Precalculus Examples

$z^{3} = - 343$

Step 1

Add $343$ to both sides of the equation.

$z^{3} + 343 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Rewrite $343$ as $7^{3}$ .

$z^{3} + 7^{3} = 0$

Step 2.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = z$ and $b = 7$ .

$(z + 7) (z^{2} - z \cdot 7 + 7^{2}) = 0$

Step 2.3

Simplify.

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Step 2.3.1

Multiply $7$ by $- 1$ .

$(z + 7) (z^{2} - 7 z + 7^{2}) = 0$

Step 2.3.2

Raise $7$ to the power of $2$ .

$(z + 7) (z^{2} - 7 z + 49) = 0$

Step 3

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

$z + 7 = 0$

$z^{2} - 7 z + 49 = 0$

Step 4

Set $z + 7$ equal to $0$ and solve for $z$ .

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Step 4.1

Set $z + 7$ equal to $0$ .

$z + 7 = 0$

Step 4.2

Subtract $7$ from both sides of the equation.

$z = - 7$

Step 5

Set $z^{2} - 7 z + 49$ equal to $0$ and solve for $z$ .

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Step 5.1

Set $z^{2} - 7 z + 49$ equal to $0$ .

$z^{2} - 7 z + 49 = 0$

Step 5.2

Solve $z^{2} - 7 z + 49 = 0$ for $z$ .

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Step 5.2.1

Use the quadratic formula to find the solutions.

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

Step 5.2.2

Substitute the values $a = 1$ , $b = - 7$ , and $c = 49$ into the quadratic formula and solve for $z$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (1 \cdot 49)}}{2 \cdot 1}$

Step 5.2.3

Simplify.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

Raise $- 7$ to the power of $2$ .

$z = \frac{7 \pm \sqrt{49 - 4 \cdot 1 \cdot 49}}{2 \cdot 1}$

Step 5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 49$ .

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Step 5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$z = \frac{7 \pm \sqrt{49 - 4 \cdot 49}}{2 \cdot 1}$

Step 5.2.3.1.2.2

Multiply $- 4$ by $49$ .

$z = \frac{7 \pm \sqrt{49 - 196}}{2 \cdot 1}$

Step 5.2.3.1.3

Subtract $196$ from $49$ .

$z = \frac{7 \pm \sqrt{- 147}}{2 \cdot 1}$

Step 5.2.3.1.4

Rewrite $- 147$ as $- 1 (147)$ .

$z = \frac{7 \pm \sqrt{- 1 \cdot 147}}{2 \cdot 1}$

Step 5.2.3.1.5

Rewrite $\sqrt{- 1 (147)}$ as $\sqrt{- 1} \cdot \sqrt{147}$ .

$z = \frac{7 \pm \sqrt{- 1} \cdot \sqrt{147}}{2 \cdot 1}$

Step 5.2.3.1.6

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

$z = \frac{7 \pm i \cdot \sqrt{147}}{2 \cdot 1}$

Step 5.2.3.1.7

Rewrite $147$ as $7^{2} \cdot 3$ .

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Step 5.2.3.1.7.1

Factor $49$ out of $147$ .

$z = \frac{7 \pm i \cdot \sqrt{49 (3)}}{2 \cdot 1}$

Step 5.2.3.1.7.2

Rewrite $49$ as $7^{2}$ .

$z = \frac{7 \pm i \cdot \sqrt{7^{2} \cdot 3}}{2 \cdot 1}$

Step 5.2.3.1.8

Pull terms out from under the radical.

$z = \frac{7 \pm i \cdot (7 \sqrt{3})}{2 \cdot 1}$

Step 5.2.3.1.9

Move $7$ to the left of $i$ .

$z = \frac{7 \pm 7 i \sqrt{3}}{2 \cdot 1}$

Step 5.2.3.2

Multiply $2$ by $1$ .

$z = \frac{7 \pm 7 i \sqrt{3}}{2}$

Step 5.2.4

The final answer is the combination of both solutions.

$z = \frac{7 + 7 i \sqrt{3}}{2}, \frac{7 - 7 i \sqrt{3}}{2}$

Step 6

The final solution is all the values that make $(z + 7) (z^{2} - 7 z + 49) = 0$ true.

$z = - 7, \frac{7 + 7 i \sqrt{3}}{2}, \frac{7 - 7 i \sqrt{3}}{2}$