Solve sqrt[3]{n-1}=sqrt{u-1}

$$\sqrt[ 3 ] { n - 1 } = \sqrt { u - 1 }$$

Solve for u

$$u=\left(n-1\right)^{\frac{2}{3}}+1$$

$$n\geq 1$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$\sqrt{u-1}=\sqrt[3]{n-1}$$

Square both sides of the equation.

$$u-1=\left(n-1\right)^{\frac{2}{3}}$$

Add 1 to both sides of the equation.

$$u-1-\left(-1\right)=\left(n-1\right)^{\frac{2}{3}}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$u=\left(n-1\right)^{\frac{2}{3}}-\left(-1\right)$$

Subtract -1 from \left(n-1\right)^{\frac{2}{3}}.

$$u=\left(n-1\right)^{\frac{2}{3}}+1$$

Solve for n

$$n=\sqrt{u-1}u-\sqrt{u-1}+1$$

$$u\geq 1$$