Solve for x cube root of 2x^3-1=x

$\sqrt[3]{2 x^{3} - 1} = x$

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{2 x^{3} - 1}}^{3} = x^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2 x^{3} - 1}$ as ${(2 x^{3} - 1)}^{\frac{1}{3}}$ .

${({(2 x^{3} - 1)}^{\frac{1}{3}})}^{3} = x^{3}$

Simplify the left side.

Simplify ${({(2 x^{3} - 1)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(2 x^{3} - 1)}^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(2 x^{3} - 1)}^{\frac{1}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(2 x^{3} - 1)}^{\frac{1}{3} \cdot 3} = x^{3}$

Rewrite the expression.

${(2 x^{3} - 1)}^{1} = x^{3}$

Simplify.

$2 x^{3} - 1 = x^{3}$

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{3}$ from both sides of the equation.

$2 x^{3} - 1 - x^{3} = 0$

Subtract $x^{3}$ from $2 x^{3}$ .

$x^{3} - 1 = 0$

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 $- 4$ 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 cube root of 2x^3-1=x

Related Questions

Support LCMGCF.com by disabling your adblocker.