Find the Roots (Zeros) 1/6x^3=-36

$\frac{1}{6} x^{3} = - 36$

Multiply both sides of the equation by $6$ .

$6 (\frac{1}{6} x^{3}) = 6 \cdot - 36$

Simplify both sides of the equation.

Simplify the left side.

Simplify $6 (\frac{1}{6} x^{3})$ .

Combine $\frac{1}{6}$ and $x^{3}$ .

$6 \frac{x^{3}}{6} = 6 \cdot - 36$

Cancel the common factor of $6$ .

Cancel the common factor.

$6 \frac{x^{3}}{6} = 6 \cdot - 36$

Rewrite the expression.

$x^{3} = 6 \cdot - 36$

Simplify the right side.

Multiply $6$ by $- 36$ .

$x^{3} = - 216$

Add $216$ to both sides of the equation.

$x^{3} + 216 = 0$

Factor the left side of the equation.

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

$x^{3} + 6^{3} = 0$

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 $a = x$ and $b = 6$ .

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

Simplify.

Multiply $6$ by $- 1$ .

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

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

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

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

$x + 6 = 0$

$x^{2} - 6 x + 36 = 0$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

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

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

$x^{2} - 6 x + 36 = 0$

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $36$ .

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

Subtract $144$ from $36$ .

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

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

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

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

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

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

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

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

Factor $36$ out of $108$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

$x = 3 \pm 3 i \sqrt{3}$

The final answer is the combination of both solutions.

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

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

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

Find the Roots (Zeros) 1/6x^3=-36

Related Questions

Support LCMGCF.com by disabling your adblocker.