Solve for y -170=6y^3-8

$- 170 = 6 y^{3} - 8$

Rewrite the equation as $6 y^{3} - 8 = - 170$ .

$6 y^{3} - 8 = - 170$

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

Add $8$ to both sides of the equation.

$6 y^{3} = - 170 + 8$

Add $- 170$ and $8$ .

$6 y^{3} = - 162$

Add $162$ to both sides of the equation.

$6 y^{3} + 162 = 0$

Factor the left side of the equation.

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

Factor $6$ out of $6 y^{3}$ .

$6 (y^{3}) + 162 = 0$

Factor $6$ out of $162$ .

$6 y^{3} + 6 \cdot 27 = 0$

Factor $6$ out of $6 y^{3} + 6 \cdot 27$ .

$6 (y^{3} + 27) = 0$

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

$6 (y^{3} + 3^{3}) = 0$

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

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

Factor.

Simplify.

Multiply $3$ by $- 1$ .

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

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

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

Remove unnecessary parentheses.

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

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

$y + 3 = 0$

$y^{2} - 3 y + 9 = 0$

Set $y + 3$ equal to $0$ and solve for $y$ .

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

Subtract $3$ from both sides of the equation.

$y = - 3$

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

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

$y^{2} - 3 y + 9 = 0$

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

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 $y$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $9$ .

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

Subtract $36$ from $9$ .

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

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

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

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

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

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

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

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

Factor $9$ out of $27$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

$y = \frac{3 \pm 3 i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

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

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

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

Solve for y -170=6y^3-8

Related Questions

Support LCMGCF.com by disabling your adblocker.