Solve by Factoring 8z^3-1=0

$8 z^{3} - 1 = 0$

Step 1

Rewrite $8 z^{3}$ as ${(2 z)}^{3}$ .

${(2 z)}^{3} - 1 = 0$

Step 2

Rewrite $1$ as $1^{3}$ .

${(2 z)}^{3} - 1^{3} = 0$

Step 3

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 $a = 2 z$ and $b = 1$ .

$(2 z - 1) ({(2 z)}^{2} + 2 z \cdot 1 + 1^{2}) = 0$

Step 4

Simplify.

Apply the product rule to $2 z$ .

$(2 z - 1) (2^{2} z^{2} + 2 z \cdot 1 + 1^{2}) = 0$

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

$(2 z - 1) (4 z^{2} + 2 z \cdot 1 + 1^{2}) = 0$

Multiply $2$ by $1$ .

$(2 z - 1) (4 z^{2} + 2 z + 1^{2}) = 0$

One to any power is one.

$(2 z - 1) (4 z^{2} + 2 z + 1) = 0$

Step 5

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

$2 z - 1 = 0$

$4 z^{2} + 2 z + 1 = 0$

Step 6

Set $2 z - 1$ equal to $0$ and solve for $z$ .

Set $2 z - 1$ equal to $0$ .

$2 z - 1 = 0$

Solve $2 z - 1 = 0$ for $z$ .

Add $1$ to both sides of the equation.

$2 z = 1$

Divide each term in $2 z = 1$ by $2$ and simplify.

Divide each term in $2 z = 1$ by $2$ .

$\frac{2 z}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 z}{2} = \frac{1}{2}$

Divide $z$ by $1$ .

$z = \frac{1}{2}$

Step 7

Set $4 z^{2} + 2 z + 1$ equal to $0$ and solve for $z$ .

Set $4 z^{2} + 2 z + 1$ equal to $0$ .

$4 z^{2} + 2 z + 1 = 0$

Solve $4 z^{2} + 2 z + 1 = 0$ for $z$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = 2$ , and $c = 1$ into the quadratic formula and solve for $z$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$z = \frac{- 2 \pm \sqrt{4 - 16 \cdot 1}}{2 \cdot 4}$

Multiply $- 16$ by $1$ .

$z = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

Subtract $16$ from $4$ .

$z = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

Rewrite $- 12$ as $- 1 (12)$ .

$z = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$z = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

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

$z = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$z = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$z = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$z = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Move $2$ to the left of $i$ .

$z = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$z = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$z = \frac{- 1 \pm i \sqrt{3}}{4}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$z = \frac{- 2 \pm \sqrt{4 - 16 \cdot 1}}{2 \cdot 4}$

Multiply $- 16$ by $1$ .

$z = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

Subtract $16$ from $4$ .

$z = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

Rewrite $- 12$ as $- 1 (12)$ .

$z = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$z = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

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

$z = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$z = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$z = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$z = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Move $2$ to the left of $i$ .

$z = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$z = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$z = \frac{- 1 \pm i \sqrt{3}}{4}$

Change the $\pm$ to $+$ .

$z = \frac{- 1 + i \sqrt{3}}{4}$

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

$z = \frac{- 1 \cdot 1 + i \sqrt{3}}{4}$

Factor $- 1$ out of $i \sqrt{3}$ .

$z = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{4}$

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$z = \frac{- 1 (1 - i \sqrt{3})}{4}$

Move the negative in front of the fraction.

$z = - \frac{1 - i \sqrt{3}}{4}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$z = \frac{- 2 \pm \sqrt{4 - 16 \cdot 1}}{2 \cdot 4}$

Multiply $- 16$ by $1$ .

$z = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

Subtract $16$ from $4$ .

$z = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

Rewrite $- 12$ as $- 1 (12)$ .

$z = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$z = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

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

$z = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$z = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$z = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$z = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Move $2$ to the left of $i$ .

$z = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$z = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$z = \frac{- 1 \pm i \sqrt{3}}{4}$

Change the $\pm$ to $-$ .

$z = \frac{- 1 - i \sqrt{3}}{4}$

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

$z = \frac{- 1 \cdot 1 - i \sqrt{3}}{4}$

Factor $- 1$ out of $- i \sqrt{3}$ .

$z = \frac{- 1 \cdot 1 - (i \sqrt{3})}{4}$

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$z = \frac{- 1 (1 + i \sqrt{3})}{4}$

Move the negative in front of the fraction.

$z = - \frac{1 + i \sqrt{3}}{4}$

The final answer is the combination of both solutions.

$z = - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$

Step 8

The final solution is all the values that make $(2 z - 1) (4 z^{2} + 2 z + 1) = 0$ true.

$z = \frac{1}{2}, - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$

Solve by Factoring 8z^3-1=0

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