Basic Math Examples

$z = \frac{1}{z^{2}}$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, z^{2}$

Step 1.2

The LCM of one and any expression is the expression.

$z^{2}$

Step 2

Multiply each term in $z = \frac{1}{z^{2}}$ by $z^{2}$ to eliminate the fractions.

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Step 2.1

Multiply each term in $z = \frac{1}{z^{2}}$ by $z^{2}$ .

$z \cdot z^{2} = \frac{1}{z^{2}} z^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Multiply $z$ by $z^{2}$ by adding the exponents.

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Step 2.2.1.1

Multiply $z$ by $z^{2}$ .

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Step 2.2.1.1.1

Raise $z$ to the power of $1$ .

$z^{1} z^{2} = \frac{1}{z^{2}} z^{2}$

Step 2.2.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$z^{1 + 2} = \frac{1}{z^{2}} z^{2}$

Step 2.2.1.2

Add $1$ and $2$ .

$z^{3} = \frac{1}{z^{2}} z^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $z^{2}$ .

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Step 2.3.1.1

Cancel the common factor.

$z^{3} = \frac{1}{z^{2}} z^{2}$

Step 2.3.1.2

Rewrite the expression.

$z^{3} = 1$

Step 3

Solve the equation.

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Step 3.1

Subtract $1$ from both sides of the equation.

$z^{3} - 1 = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

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

$z^{3} - 1^{3} = 0$

Step 3.2.2

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 $a = z$ and $b = 1$ .

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

Step 3.2.3

Simplify.

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Step 3.2.3.1

Multiply $z$ by $1$ .

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

Step 3.2.3.2

One to any power is one.

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

Step 3.3

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

$z - 1 = 0$

$z^{2} + z + 1 = 0$

Step 3.4

Set $z - 1$ equal to $0$ and solve for $z$ .

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Step 3.4.1

Set $z - 1$ equal to $0$ .

$z - 1 = 0$

Step 3.4.2

Add $1$ to both sides of the equation.

$z = 1$

Step 3.5

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

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Step 3.5.1

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

$z^{2} + z + 1 = 0$

Step 3.5.2

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

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Step 3.5.2.1

Use the quadratic formula to find the solutions.

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

Step 3.5.2.2

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

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

Step 3.5.2.3

Simplify.

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Step 3.5.2.3.1

Simplify the numerator.

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Step 3.5.2.3.1.1

One to any power is one.

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

Step 3.5.2.3.1.2

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

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Step 3.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 3.5.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 3.5.2.3.1.3

Subtract $4$ from $1$ .

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

Step 3.5.2.3.1.4

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

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

Step 3.5.2.3.1.5

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

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

Step 3.5.2.3.1.6

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

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

Step 3.5.2.3.2

Multiply $2$ by $1$ .

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

Step 3.5.2.4

The final answer is the combination of both solutions.

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

Step 3.6

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

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