Solve by Factoring 8x^3+1=0

$8 x^{3} + 1 = 0$

Step 1

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

${(2 x)}^{3} + 1 = 0$

Step 2

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

${(2 x)}^{3} + 1^{3} = 0$

Step 3

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

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

Step 4

Simplify.

Apply the product rule to $2 x$ .

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

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

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

Multiply $2$ by $- 1$ .

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

Multiply $- 2$ by $1$ .

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

One to any power is one.

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

Step 5

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

$2 x + 1 = 0$

$4 x^{2} - 2 x + 1 = 0$

Step 6

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

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

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 7

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

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

$4 x^{2} - 2 x + 1 = 0$

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

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 $x$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $1$ .

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

Subtract $16$ from $4$ .

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

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

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

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

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

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

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

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

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $4$ .

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

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

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $1$ .

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

Subtract $16$ from $4$ .

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

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

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

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

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

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

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

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

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $4$ .

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

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

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

Change the $\pm$ to $+$ .

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $1$ .

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

Subtract $16$ from $4$ .

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

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

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

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

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

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

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

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

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $4$ .

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

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

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

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

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

Step 8

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

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

Solve by Factoring 8x^3+1=0

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