Solve the Rational Equation for x cube root of x^2=9

$\sqrt[3]{x^{2}} = 9$

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

${\sqrt[3]{x^{2}}}^{3} = 9^{3}$

Simplify each side of the equation.

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

${(x^{\frac{2}{3}})}^{3} = 9^{3}$

Simplify the left side.

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

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

$x^{\frac{2}{3} \cdot 3} = 9^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x^{\frac{2}{3} \cdot 3} = 9^{3}$

Rewrite the expression.

$x^{2} = 9^{3}$

Simplify the right side.

Raise $9$ to the power of $3$ .

$x^{2} = 729$

Solve for $x$ .

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

$x = \pm \sqrt{729}$

Simplify $\pm \sqrt{729}$ .

Rewrite $729$ as $27^{2}$ .

$x = \pm \sqrt{27^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 27$

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.

$x = 27$

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

$x = - 27$

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

$x = 27, - 27$