Solve by Factoring cube root of x-4-2=0

$\sqrt[3]{x - 4} - 2 = 0$

Step 1

Add $2$ to both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$x - 4 = 2^{3}$

Simplify the right side.

Raise $2$ to the power of $3$ .

$x - 4 = 8$

Step 4

Move all terms not containing $x$ to the right side of the equation.

Add $4$ to both sides of the equation.

$x = 8 + 4$

Add $8$ and $4$ .

$x = 12$