Solve for x cube root of x-1=0

$\sqrt[3]{x - 1} = 0$

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

${\sqrt[3]{x - 1}}^{3} = 0^{3}$

Simplify each side of the equation.

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

${({(x - 1)}^{\frac{1}{3}})}^{3} = 0^{3}$

Simplify the left side.

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

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

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

${(x - 1)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x - 1)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Rewrite the expression.

${(x - 1)}^{1} = 0^{3}$

Simplify.

$x - 1 = 0^{3}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$