Solve for x cube root of x^2-12 = cube root of 4x

$\sqrt[3]{x^{2} - 12} = \sqrt[3]{4 x}$

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

${\sqrt[3]{x^{2} - 12}}^{3} = {\sqrt[3]{4 x}}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x^{2} - 12}$ as ${(x^{2} - 12)}^{\frac{1}{3}}$ .

${({(x^{2} - 12)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{4 x}}^{3}$

Simplify the left side.

Simplify ${({(x^{2} - 12)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(x^{2} - 12)}^{\frac{1}{3}})}^{3}$ .

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

${(x^{2} - 12)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{4 x}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{2} - 12)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{4 x}}^{3}$

Rewrite the expression.

${(x^{2} - 12)}^{1} = {\sqrt[3]{4 x}}^{3}$

Simplify.

$x^{2} - 12 = {\sqrt[3]{4 x}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{4 x}}^{3}$ as $4 x$ .

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

$x^{2} - 12 = {({(4 x)}^{\frac{1}{3}})}^{3}$

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

$x^{2} - 12 = {(4 x)}^{\frac{1}{3} \cdot 3}$

Combine $\frac{1}{3}$ and $3$ .

$x^{2} - 12 = {(4 x)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x^{2} - 12 = {(4 x)}^{\frac{3}{3}}$

Rewrite the expression.

$x^{2} - 12 = {(4 x)}^{1}$

Simplify.

$x^{2} - 12 = 4 x$

Solve for $x$ .

Subtract $4 x$ from both sides of the equation.

$x^{2} - 12 - 4 x = 0$

Factor $x^{2} - 12 - 4 x$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$(x - 6) (x + 2) = 0$

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

$x - 6 = 0$

$x + 2 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

Solve for x cube root of x^2-12 = cube root of 4x

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