Solve for r r^3+1=0

$r^{3} + 1 = 0$

Step 1

Subtract $1$ from both sides of the equation.

$r^{3} = - 1$

Step 2

Add $1$ to both sides of the equation.

$r^{3} + 1 = 0$

Step 3

Factor the left side of the equation.

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

$r^{3} + 1^{3} = 0$

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 $a = r$ and $b = 1$ .

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

Simplify.

Multiply $- 1$ by $1$ .

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

One to any power is one.

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

Step 4

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

$r + 1 = 0$

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

Step 5

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

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

$r + 1 = 0$

Subtract $1$ from both sides of the equation.

$r = - 1$

Step 6

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

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

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

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

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 $r$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

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

Step 7

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

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

Solve for r r^3+1=0

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