Solve for j cube root of j=6

$\sqrt[3]{j} = 6$

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

${\sqrt[3]{j}}^{3} = 6^{3}$

Simplify each side of the equation.

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

${(j^{\frac{1}{3}})}^{3} = 6^{3}$

Simplify the left side.

Simplify ${(j^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${(j^{\frac{1}{3}})}^{3}$ .

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

$j^{\frac{1}{3} \cdot 3} = 6^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$j^{\frac{1}{3} \cdot 3} = 6^{3}$

Rewrite the expression.

$j^{1} = 6^{3}$

Simplify.

$j = 6^{3}$

Simplify the right side.

Raise $6$ to the power of $3$ .

$j = 216$