Solve for y -10y^3-5=5

$- 10 y^{3} - 5 = 5$

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

Add $5$ to both sides of the equation.

$- 10 y^{3} = 5 + 5$

Add $5$ and $5$ .

$- 10 y^{3} = 10$

Subtract $10$ from both sides of the equation.

$- 10 y^{3} - 10 = 0$

Factor the left side of the equation.

Factor $- 10$ out of $- 10 y^{3} - 10$ .

Factor $- 10$ out of $- 10 y^{3}$ .

$- 10 y^{3} - 10 = 0$

Factor $- 10$ out of $- 10$ .

$- 10 y^{3} - 10 \cdot 1 = 0$

Factor $- 10$ out of $- 10 (y^{3}) - 10 (1)$ .

$- 10 (y^{3} + 1) = 0$

Rewrite $1$ as $1^{3}$ .

$- 10 (y^{3} + 1^{3}) = 0$

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 $b = 1$ .

$- 10 ((y + 1) (y^{2} - y \cdot 1 + 1^{2})) = 0$

Factor.

Simplify.

Multiply $- 1$ by $1$ .

$- 10 ((y + 1) (y^{2} - y + 1^{2})) = 0$

One to any power is one.

$- 10 ((y + 1) (y^{2} - y + 1)) = 0$

Remove unnecessary parentheses.

$- 10 (y + 1) (y^{2} - y + 1) = 0$

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

$y + 1 = 0$

$y^{2} - y + 1 = 0$

Set $y + 1$ equal to $0$ and solve for $y$ .

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Subtract $1$ from both sides of the equation.

$y = - 1$

Set $y^{2} - y + 1$ equal to $0$ and solve for $y$ .

Set $y^{2} - y + 1$ equal to $0$ .

$y^{2} - y + 1 = 0$

Solve $y^{2} - y + 1 = 0$ for $y$ .

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

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

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

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

$y = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

The final solution is all the values that make $- 10 (y + 1) (y^{2} - y + 1) = 0$ true.

$y = - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Solve for y -10y^3-5=5

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