Solve for x cube root of 4x^2+2x-8=x

$\sqrt[3]{4 x^{2} + 2 x - 8} = x$

Step 1

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

${\sqrt[3]{4 x^{2} + 2 x - 8}}^{3} = x^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 x^{2} + 2 x - 8}$ as ${(4 x^{2} + 2 x - 8)}^{\frac{1}{3}}$ .

${({(4 x^{2} + 2 x - 8)}^{\frac{1}{3}})}^{3} = x^{3}$

Simplify the left side.

Simplify ${({(4 x^{2} + 2 x - 8)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(4 x^{2} + 2 x - 8)}^{\frac{1}{3}})}^{3}$ .

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

${(4 x^{2} + 2 x - 8)}^{\frac{1}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(4 x^{2} + 2 x - 8)}^{\frac{1}{3} \cdot 3} = x^{3}$

Rewrite the expression.

${(4 x^{2} + 2 x - 8)}^{1} = x^{3}$

Simplify.

$4 x^{2} + 2 x - 8 = x^{3}$

Step 3

Solve for $x$ .

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

$4 x^{2} + 2 x - 8 - x^{3} = 0$

Factor the left side of the equation.

Reorder terms.

$- x^{3} + 4 x^{2} + 2 x - 8 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(- x^{3} + 4 x^{2}) + 2 x - 8 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- x + 4$ .

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

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

$- x + 4 = 0$

$x^{2} - 2 = 0$

Set $- x + 4$ equal to $0$ and solve for $x$ .

Set $- x + 4$ equal to $0$ .

$- x + 4 = 0$

Solve $- x + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- x = - 4$

Divide each term in $- x = - 4$ by $- 1$ and simplify.

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x = 4$

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

Set $x^{2} - 2$ equal to $0$ .

$x^{2} - 2 = 0$

Solve $x^{2} - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$x^{2} = 2$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{2}, - \sqrt{2}$

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

$x = 4, \sqrt{2}, - \sqrt{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = 4, \sqrt{2}, - \sqrt{2}$

Decimal Form:

$x = 4, 1.41421356 \dots, - 1.41421356 \dots$

Solve for x cube root of 4x^2+2x-8=x

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