Solve Using the Quadratic Formula 8x^3-27=0

$8 x^{3} - 27 = 0$

Step 1

Add $27$ to both sides of the equation.

$8 x^{3} = 27$

Step 2

Subtract $27$ from both sides of the equation.

$8 x^{3} - 27 = 0$

Step 3

Factor the left side of the equation.

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

${(2 x)}^{3} - 27 = 0$

Rewrite $27$ as $3^{3}$ .

${(2 x)}^{3} - 3^{3} = 0$

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 $a = 2 x$ and $b = 3$ .

$(2 x - 3) ({(2 x)}^{2} + 2 x \cdot 3 + 3^{2}) = 0$

Simplify.

Apply the product rule to $2 x$ .

$(2 x - 3) (2^{2} x^{2} + 2 x \cdot 3 + 3^{2}) = 0$

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

$(2 x - 3) (4 x^{2} + 2 x \cdot 3 + 3^{2}) = 0$

Multiply $3$ by $2$ .

$(2 x - 3) (4 x^{2} + 6 x + 3^{2}) = 0$

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

$(2 x - 3) (4 x^{2} + 6 x + 9) = 0$

Step 4

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

$2 x - 3 = 0$

$4 x^{2} + 6 x + 9 = 0$

Step 5

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 6

Set $4 x^{2} + 6 x + 9$ equal to $0$ and solve for $x$ .

Set $4 x^{2} + 6 x + 9$ equal to $0$ .

$4 x^{2} + 6 x + 9 = 0$

Solve $4 x^{2} + 6 x + 9 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = 6$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (4 \cdot 9)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 4 \cdot 9}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 9$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 6 \pm \sqrt{36 - 16 \cdot 9}}{2 \cdot 4}$

Multiply $- 16$ by $9$ .

$x = \frac{- 6 \pm \sqrt{36 - 144}}{2 \cdot 4}$

Subtract $144$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 108}}{2 \cdot 4}$

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

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 108}}{2 \cdot 4}$

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

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{108}}{2 \cdot 4}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{108}}{2 \cdot 4}$

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

Factor $36$ out of $108$ .

$x = \frac{- 6 \pm i \cdot \sqrt{36 (3)}}{2 \cdot 4}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{6^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm i \cdot (6 \sqrt{3})}{2 \cdot 4}$

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

$x = \frac{- 6 \pm 6 i \sqrt{3}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 6 \pm 6 i \sqrt{3}}{8}$

Simplify $\frac{- 6 \pm 6 i \sqrt{3}}{8}$ .

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

The final answer is the combination of both solutions.

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

Step 7

The final solution is all the values that make $(2 x - 3) (4 x^{2} + 6 x + 9) = 0$ true.

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

Solve Using the Quadratic Formula 8x^3-27=0

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