Solve for x x^2-1/x=0

$x^{2} - \frac{1}{x} = 0$

Step 1

Find the LCD of the terms in the equation.

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

$1, x, 1$

The LCM of one and any expression is the expression.

$x$

Step 2

Multiply each term in $x^{2} - \frac{1}{x} = 0$ by $x$ to eliminate the fractions.

Multiply each term in $x^{2} - \frac{1}{x} = 0$ by $x$ .

$x^{2} x - \frac{1}{x} x = 0 x$

Simplify the left side.

Simplify each term.

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

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

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

$x^{2} x^{1} - \frac{1}{x} x = 0 x$

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

$x^{2 + 1} - \frac{1}{x} x = 0 x$

Add $2$ and $1$ .

$x^{3} - \frac{1}{x} x = 0 x$

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{1}{x}$ into the numerator.

$x^{3} + \frac{- 1}{x} x = 0 x$

Cancel the common factor.

$x^{3} + \frac{- 1}{x} x = 0 x$

Rewrite the expression.

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

Simplify the right side.

Multiply $0$ by $x$ .

$x^{3} - 1 = 0$

Step 3

Solve the equation.

Add $1$ to both sides of the equation.

$x^{3} = 1$

Subtract $1$ from both sides of the equation.

$x^{3} - 1 = 0$

Factor the left side of the equation.

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

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

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 $a = x$ and $b = 1$ .

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

Simplify.

Multiply $x$ by $1$ .

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

One to any power is one.

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

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

$x - 1 = 0$

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

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

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

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

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 $x$ .

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

Simplify.

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

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

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

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

Solve for x x^2-1/x=0

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