Solve for b -165=-3-6b^3

$- 165 = - 3 - 6 b^{3}$

Rewrite the equation as $- 3 - 6 b^{3} = - 165$ .

$- 3 - 6 b^{3} = - 165$

Move all terms not containing $b$ to the right side of the equation.

Add $3$ to both sides of the equation.

$- 6 b^{3} = - 165 + 3$

Add $- 165$ and $3$ .

$- 6 b^{3} = - 162$

Add $162$ to both sides of the equation.

$- 6 b^{3} + 162 = 0$

Factor the left side of the equation.

Factor $- 6$ out of $- 6 b^{3} + 162$ .

Factor $- 6$ out of $- 6 b^{3}$ .

$- 6 b^{3} + 162 = 0$

Factor $- 6$ out of $162$ .

$- 6 b^{3} - 6 \cdot - 27 = 0$

Factor $- 6$ out of $- 6 (b^{3}) - 6 (- 27)$ .

$- 6 (b^{3} - 27) = 0$

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

$- 6 (b^{3} - 3^{3}) = 0$

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 $a = b$ and $b = 3$ .

$- 6 ((b - 3) (b^{2} + b \cdot 3 + 3^{2})) = 0$

Factor.

Simplify.

Move $3$ to the left of $b$ .

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

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

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

Remove unnecessary parentheses.

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

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

$b - 3 = 0$

$b^{2} + 3 b + 9 = 0$

Set $b - 3$ equal to $0$ and solve for $b$ .

Set $b - 3$ equal to $0$ .

$b - 3 = 0$

Add $3$ to both sides of the equation.

$b = 3$

Set $b^{2} + 3 b + 9$ equal to $0$ and solve for $b$ .

Set $b^{2} + 3 b + 9$ equal to $0$ .

$b^{2} + 3 b + 9 = 0$

Solve $b^{2} + 3 b + 9 = 0$ for $b$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 3$ , and $c = 9$ into the quadratic formula and solve for $b$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$b = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$b = \frac{- 3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$b = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$b = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$b = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$b = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$b = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$b = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical.

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

Move $3$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

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

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

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

Solve for b -165=-3-6b^3

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