Solve for n cube root of n-9=3

$\sqrt[3]{n - 9} = 3$

Step 1

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

${\sqrt[3]{n - 9}}^{3} = 3^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{n - 9}$ as ${(n - 9)}^{\frac{1}{3}}$ .

${({(n - 9)}^{\frac{1}{3}})}^{3} = 3^{3}$

Simplify the left side.

Simplify ${({(n - 9)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(n - 9)}^{\frac{1}{3}})}^{3}$ .

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

${(n - 9)}^{\frac{1}{3} \cdot 3} = 3^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(n - 9)}^{\frac{1}{3} \cdot 3} = 3^{3}$

Rewrite the expression.

${(n - 9)}^{1} = 3^{3}$

Simplify.

$n - 9 = 3^{3}$

Simplify the right side.

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

$n - 9 = 27$

Step 3

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

Add $9$ to both sides of the equation.

$n = 27 + 9$

Add $27$ and $9$ .

$n = 36$