Basic Math Examples

$- \frac{1}{27} = r^{3}$

Step 1

Rewrite the equation as $r^{3} = - \frac{1}{27}$ .

$r^{3} = - \frac{1}{27}$

Step 2

Add $\frac{1}{27}$ to both sides of the equation.

$r^{3} + \frac{1}{27} = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

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

$r^{3} + \frac{1^{3}}{27} = 0$

Step 3.2

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

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

Step 3.3

Rewrite $\frac{1^{3}}{3^{3}}$ as ${(\frac{1}{3})}^{3}$ .

$r^{3} + {(\frac{1}{3})}^{3} = 0$

Step 3.4

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 $b = \frac{1}{3}$ .

$(r + \frac{1}{3}) (r^{2} - r \frac{1}{3} + {(\frac{1}{3})}^{2}) = 0$

Step 3.5

Simplify.

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Step 3.5.1

Combine $\frac{1}{3}$ and $r$ .

$(r + \frac{1}{3}) (r^{2} - \frac{r}{3} + {(\frac{1}{3})}^{2}) = 0$

Step 3.5.2

Apply the product rule to $\frac{1}{3}$ .

$(r + \frac{1}{3}) (r^{2} - \frac{r}{3} + \frac{1^{2}}{3^{2}}) = 0$

Step 3.5.3

One to any power is one.

$(r + \frac{1}{3}) (r^{2} - \frac{r}{3} + \frac{1}{3^{2}}) = 0$

Step 3.5.4

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

$(r + \frac{1}{3}) (r^{2} - \frac{r}{3} + \frac{1}{9}) = 0$

Step 4

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

$r + \frac{1}{3} = 0$

$r^{2} - \frac{r}{3} + \frac{1}{9} = 0$

Step 5

Set $r + \frac{1}{3}$ equal to $0$ and solve for $r$ .

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Step 5.1

Set $r + \frac{1}{3}$ equal to $0$ .

$r + \frac{1}{3} = 0$

Step 5.2

Subtract $\frac{1}{3}$ from both sides of the equation.

$r = - \frac{1}{3}$

Step 6

Set $r^{2} - \frac{r}{3} + \frac{1}{9}$ equal to $0$ and solve for $r$ .

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Step 6.1

Set $r^{2} - \frac{r}{3} + \frac{1}{9}$ equal to $0$ .

$r^{2} - \frac{r}{3} + \frac{1}{9} = 0$

Step 6.2

Solve $r^{2} - \frac{r}{3} + \frac{1}{9} = 0$ for $r$ .

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Step 6.2.1

Multiply through by the least common denominator $9$ , then simplify.

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Step 6.2.1.1

Apply the distributive property.

$9 r^{2} + 9 (- \frac{r}{3}) + 9 (\frac{1}{9}) = 0$

Step 6.2.1.2

Simplify.

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Step 6.2.1.2.1

Cancel the common factor of $3$ .

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Step 6.2.1.2.1.1

Move the leading negative in $- \frac{r}{3}$ into the numerator.

$9 r^{2} + 9 (\frac{- r}{3}) + 9 (\frac{1}{9}) = 0$

Step 6.2.1.2.1.2

Factor $3$ out of $9$ .

$9 r^{2} + 3 (3) (\frac{- r}{3}) + 9 (\frac{1}{9}) = 0$

Step 6.2.1.2.1.3

Cancel the common factor.

$9 r^{2} + 3 \cdot (3 (\frac{- r}{3})) + 9 (\frac{1}{9}) = 0$

Step 6.2.1.2.1.4

Rewrite the expression.

$9 r^{2} + 3 (- r) + 9 (\frac{1}{9}) = 0$

Step 6.2.1.2.2

Multiply $- 1$ by $3$ .

$9 r^{2} - 3 r + 9 (\frac{1}{9}) = 0$

Step 6.2.1.2.3

Cancel the common factor of $9$ .

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Step 6.2.1.2.3.1

Cancel the common factor.

$9 r^{2} - 3 r + 9 (\frac{1}{9}) = 0$

Step 6.2.1.2.3.2

Rewrite the expression.

$9 r^{2} - 3 r + 1 = 0$

Step 6.2.2

Use the quadratic formula to find the solutions.

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

Step 6.2.3

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

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

Step 6.2.4

Simplify.

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Step 6.2.4.1

Simplify the numerator.

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Step 6.2.4.1.1

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

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

Step 6.2.4.1.2

Multiply $- 4 \cdot 9 \cdot 1$ .

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Step 6.2.4.1.2.1

Multiply $- 4$ by $9$ .

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

Step 6.2.4.1.2.2

Multiply $- 36$ by $1$ .

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

Step 6.2.4.1.3

Subtract $36$ from $9$ .

$r = \frac{3 \pm \sqrt{- 27}}{2 \cdot 9}$

Step 6.2.4.1.4

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

$r = \frac{3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 9}$

Step 6.2.4.1.5

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

$r = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 9}$

Step 6.2.4.1.6

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

$r = \frac{3 \pm i \cdot \sqrt{27}}{2 \cdot 9}$

Step 6.2.4.1.7

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

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Step 6.2.4.1.7.1

Factor $9$ out of $27$ .

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

Step 6.2.4.1.7.2

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

$r = \frac{3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 9}$

Step 6.2.4.1.8

Pull terms out from under the radical.

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

Step 6.2.4.1.9

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

$r = \frac{3 \pm 3 i \sqrt{3}}{2 \cdot 9}$

Step 6.2.4.2

Multiply $2$ by $9$ .

$r = \frac{3 \pm 3 i \sqrt{3}}{18}$

Step 6.2.4.3

Simplify $\frac{3 \pm 3 i \sqrt{3}}{18}$ .

$r = \frac{1 \pm i \sqrt{3}}{6}$

Step 6.2.5

The final answer is the combination of both solutions.

$r = \frac{1 + i \sqrt{3}}{6}, \frac{1 - i \sqrt{3}}{6}$

Step 7

The final solution is all the values that make $(r + \frac{1}{3}) (r^{2} - \frac{r}{3} + \frac{1}{9}) = 0$ true.

$r = - \frac{1}{3}, \frac{1 + i \sqrt{3}}{6}, \frac{1 - i \sqrt{3}}{6}$