Solve for v 216=v^3

$216 = v^{3}$

Rewrite the equation as $v^{3} = 216$ .

$v^{3} = 216$

Subtract $216$ from both sides of the equation.

$v^{3} - 216 = 0$

Factor the left side of the equation.

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

$v^{3} - 6^{3} = 0$

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 $a = v$ and $b = 6$ .

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

Simplify.

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

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

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

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

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

$v - 6 = 0$

$v^{2} + 6 v + 36 = 0$

Set $v - 6$ equal to $0$ and solve for $v$ .

Set $v - 6$ equal to $0$ .

$v - 6 = 0$

Add $6$ to both sides of the equation.

$v = 6$

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

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

$v^{2} + 6 v + 36 = 0$

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

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 $v$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $36$ .

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

Subtract $144$ from $36$ .

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

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

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

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

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

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

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

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

Factor $36$ out of $108$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

$v = - 3 \pm 3 i \sqrt{3}$

The final answer is the combination of both solutions.

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

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

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

Solve for v 216=v^3

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