Solve for y 8-7y^3=1

$8 - 7 y^{3} = 1$

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

Subtract $8$ from both sides of the equation.

$- 7 y^{3} = 1 - 8$

Subtract $8$ from $1$ .

$- 7 y^{3} = - 7$

Add $7$ to both sides of the equation.

$- 7 y^{3} + 7 = 0$

Factor the left side of the equation.

Factor $- 7$ out of $- 7 y^{3} + 7$ .

Factor $- 7$ out of $- 7 y^{3}$ .

$- 7 y^{3} + 7 = 0$

Factor $- 7$ out of $7$ .

$- 7 y^{3} - 7 \cdot - 1 = 0$

Factor $- 7$ out of $- 7 (y^{3}) - 7 (- 1)$ .

$- 7 (y^{3} - 1) = 0$

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

$- 7 (y^{3} - 1^{3}) = 0$

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 $a = y$ and $b = 1$ .

$- 7 ((y - 1) (y^{2} + y \cdot 1 + 1^{2})) = 0$

Factor.

Simplify.

Multiply $y$ by $1$ .

$- 7 ((y - 1) (y^{2} + y + 1^{2})) = 0$

One to any power is one.

$- 7 ((y - 1) (y^{2} + y + 1)) = 0$

Remove unnecessary parentheses.

$- 7 (y - 1) (y^{2} + y + 1) = 0$

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 $0$ .

$y - 1 = 0$

Add $1$ to 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.

One to any power is one.

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

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 $- 7 (y - 1) (y^{2} + y + 1) = 0$ true.

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

Solve for y 8-7y^3=1

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