Solve for x square root of x = cube root of x

$\sqrt{x} = \sqrt[3]{x}$

Step 1

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

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

Step 2

Simplify each side of the equation.

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

${(x^{\frac{1}{2}})}^{2} = {\sqrt[3]{x}}^{2}$

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Rewrite ${\sqrt[3]{x}}^{2}$ as ${(x^{2})}^{\frac{1}{3}}$ .

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

Step 3

Rewrite the equation as $\sqrt[3]{x^{2}} = x$ .

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

Step 4

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

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

Step 5

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

Solve for $x$ .

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

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

Factor $x^{2}$ out of $x^{2} - x^{3}$ .

Multiply by $1$ .

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

Factor $x^{2}$ out of $- x^{3}$ .

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

Factor $x^{2}$ out of $x^{2} \cdot 1 + x^{2} (- x)$ .

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

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

$x^{2} = 0$

$1 - x = 0$

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

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

$x^{2} = 0$

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

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

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

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

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

$x = 0, 1$

Solve for x square root of x = cube root of x

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