Solve for x cube root of x^3-3x^2+2=x-1

$\sqrt[3]{x^{3} - 3 x^{2} + 2} = x - 1$

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

${\sqrt[3]{x^{3} - 3 x^{2} + 2}}^{3} = {(x - 1)}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x^{3} - 3 x^{2} + 2}$ as ${(x^{3} - 3 x^{2} + 2)}^{\frac{1}{3}}$ .

${({(x^{3} - 3 x^{2} + 2)}^{\frac{1}{3}})}^{3} = {(x - 1)}^{3}$

Simplify the left side.

Simplify ${({(x^{3} - 3 x^{2} + 2)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(x^{3} - 3 x^{2} + 2)}^{\frac{1}{3}})}^{3}$ .

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

${(x^{3} - 3 x^{2} + 2)}^{\frac{1}{3} \cdot 3} = {(x - 1)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{3} - 3 x^{2} + 2)}^{\frac{1}{3} \cdot 3} = {(x - 1)}^{3}$

Rewrite the expression.

${(x^{3} - 3 x^{2} + 2)}^{1} = {(x - 1)}^{3}$

Simplify.

$x^{3} - 3 x^{2} + 2 = {(x - 1)}^{3}$

Simplify the right side.

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

Use the Binomial Theorem.

$x^{3} - 3 x^{2} + 2 = x^{3} + 3 x^{2} \cdot - 1 + 3 x {(- 1)}^{2} + {(- 1)}^{3}$

Simplify each term.

Multiply $- 1$ by $3$ .

$x^{3} - 3 x^{2} + 2 = x^{3} - 3 x^{2} + 3 x {(- 1)}^{2} + {(- 1)}^{3}$

Raise $- 1$ to the power of $2$ .

$x^{3} - 3 x^{2} + 2 = x^{3} - 3 x^{2} + 3 x \cdot 1 + {(- 1)}^{3}$

Multiply $3$ by $1$ .

$x^{3} - 3 x^{2} + 2 = x^{3} - 3 x^{2} + 3 x + {(- 1)}^{3}$

Raise $- 1$ to the power of $3$ .

$x^{3} - 3 x^{2} + 2 = x^{3} - 3 x^{2} + 3 x - 1$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{3} - 3 x^{2} + 3 x - 1 = x^{3} - 3 x^{2} + 2$

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{3}$ from both sides of the equation.

$x^{3} - 3 x^{2} + 3 x - 1 - x^{3} = - 3 x^{2} + 2$

Add $3 x^{2}$ to both sides of the equation.

$x^{3} - 3 x^{2} + 3 x - 1 - x^{3} + 3 x^{2} = 2$

Combine the opposite terms in $x^{3} - 3 x^{2} + 3 x - 1 - x^{3} + 3 x^{2}$ .

Subtract $x^{3}$ from $x^{3}$ .

$- 3 x^{2} + 3 x - 1 + 0 + 3 x^{2} = 2$

Add $- 3 x^{2} + 3 x - 1$ and $0$ .

$- 3 x^{2} + 3 x - 1 + 3 x^{2} = 2$

Add $- 3 x^{2}$ and $3 x^{2}$ .

$3 x - 1 + 0 = 2$

Add $3 x - 1$ and $0$ .

$3 x - 1 = 2$

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

Add $1$ to both sides of the equation.

$3 x = 2 + 1$

Add $2$ and $1$ .

$3 x = 3$

Divide each term in $3 x = 3$ by $3$ and simplify.

Divide each term in $3 x = 3$ by $3$ .

$\frac{3 x}{3} = \frac{3}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{3}{3}$

Divide $x$ by $1$ .

$x = \frac{3}{3}$

Simplify the right side.

Divide $3$ by $3$ .

$x = 1$

Solve for x cube root of x^3-3x^2+2=x-1

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