Solve for a cube root of a+2=4

$\sqrt[3]{a + 2} = 4$

Step 1

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

${\sqrt[3]{a + 2}}^{3} = 4^{3}$

Step 2

Simplify each side of the equation.

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

${({(a + 2)}^{\frac{1}{3}})}^{3} = 4^{3}$

Simplify the left side.

Simplify ${({(a + 2)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(a + 2)}^{\frac{1}{3}})}^{3}$ .

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

${(a + 2)}^{\frac{1}{3} \cdot 3} = 4^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(a + 2)}^{\frac{1}{3} \cdot 3} = 4^{3}$

Rewrite the expression.

${(a + 2)}^{1} = 4^{3}$

Simplify.

$a + 2 = 4^{3}$

Simplify the right side.

Raise $4$ to the power of $3$ .

$a + 2 = 64$

Step 3

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

Subtract $2$ from both sides of the equation.

$a = 64 - 2$

Subtract $2$ from $64$ .

$a = 62$