Solve for j cube root of j=6

Solve Math

Algebra

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

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.

Tap for more steps...

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.

Tap for more steps...

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

Tap for more steps...

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

Tap for more steps...

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 $3$ .

Tap for more steps...

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.

Tap for more steps...

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

$j = 216$

Do you know how to Solve for j cube root of j=6? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.

Expand using the Binomial Theorem (x-y)^2

Evaluate 20/4

Find the X and Y Intercepts -x+4y=4

Write in Standard Form (-2x^4+7x^2y-7)+(-9x^3+7xy+7)

Find the Domain y=2 square root of x