Finite Math Examples

$8 y^{3} - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$8 y^{3} = 1$

Step 2

Subtract $1$ from both sides of the equation.

$8 y^{3} - 1 = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Rewrite $8 y^{3}$ as ${(2 y)}^{3}$ .

${(2 y)}^{3} - 1 = 0$

Step 3.2

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

${(2 y)}^{3} - 1^{3} = 0$

Step 3.3

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 $a = 2 y$ and $b = 1$ .

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

Step 3.4

Simplify.

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Step 3.4.1

Apply the product rule to $2 y$ .

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

Step 3.4.2

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

$(2 y - 1) (4 y^{2} + 2 y \cdot 1 + 1^{2}) = 0$

Step 3.4.3

Multiply $2$ by $1$ .

$(2 y - 1) (4 y^{2} + 2 y + 1^{2}) = 0$

Step 3.4.4

One to any power is one.

$(2 y - 1) (4 y^{2} + 2 y + 1) = 0$

Step 4

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

$2 y - 1 = 0$

$4 y^{2} + 2 y + 1 = 0$

Step 5

Set $2 y - 1$ equal to $0$ and solve for $y$ .

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Step 5.1

Set $2 y - 1$ equal to $0$ .

$2 y - 1 = 0$

Step 5.2

Solve $2 y - 1 = 0$ for $y$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$2 y = 1$

Step 5.2.2

Divide each term in $2 y = 1$ by $2$ and simplify.

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Step 5.2.2.1

Divide each term in $2 y = 1$ by $2$ .

$\frac{2 y}{2} = \frac{1}{2}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{1}{2}$

Step 5.2.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{1}{2}$

Step 6

Set $4 y^{2} + 2 y + 1$ equal to $0$ and solve for $y$ .

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Step 6.1

Set $4 y^{2} + 2 y + 1$ equal to $0$ .

$4 y^{2} + 2 y + 1 = 0$

Step 6.2

Solve $4 y^{2} + 2 y + 1 = 0$ for $y$ .

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Step 6.2.1

Use the quadratic formula to find the solutions.

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

Step 6.2.2

Substitute the values $a = 4$ , $b = 2$ , and $c = 1$ into the quadratic formula and solve for $y$ .

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

Step 6.2.3

Simplify.

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Step 6.2.3.1

Simplify the numerator.

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Step 6.2.3.1.1

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

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

Step 6.2.3.1.2

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

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Step 6.2.3.1.2.1

Multiply $- 4$ by $4$ .

$y = \frac{- 2 \pm \sqrt{4 - 16 \cdot 1}}{2 \cdot 4}$

Step 6.2.3.1.2.2

Multiply $- 16$ by $1$ .

$y = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

Step 6.2.3.1.3

Subtract $16$ from $4$ .

$y = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

Step 6.2.3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Step 6.2.3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

Step 6.2.3.1.6

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

$y = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Step 6.2.3.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 6.2.3.1.7.1

Factor $4$ out of $12$ .

$y = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

Step 6.2.3.1.7.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Step 6.2.3.1.8

Pull terms out from under the radical.

$y = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Step 6.2.3.1.9

Move $2$ to the left of $i$ .

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

Step 6.2.3.2

Multiply $2$ by $4$ .

$y = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Step 6.2.3.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$y = \frac{- 1 \pm i \sqrt{3}}{4}$

Step 6.2.4

The final answer is the combination of both solutions.

$y = - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$

Step 7

The final solution is all the values that make $(2 y - 1) (4 y^{2} + 2 y + 1) = 0$ true.

$y = \frac{1}{2}, - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$