Find the Inverse cube root of x-1

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{y} - 1 = x$ .

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

Add $1$ to both sides of the equation.

$\sqrt[3]{y} = x + 1$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$y^{1} = {(x + 1)}^{3}$

Simplify.

$y = {(x + 1)}^{3}$

Simplify the right side.

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

Use the Binomial Theorem.

$y = x^{3} + 3 x^{2} \cdot 1 + 3 x \cdot 1^{2} + 1^{3}$

Simplify each term.

Multiply $3$ by $1$ .

$y = x^{3} + 3 x^{2} + 3 x \cdot 1^{2} + 1^{3}$

One to any power is one.

$y = x^{3} + 3 x^{2} + 3 x \cdot 1 + 1^{3}$

Multiply $3$ by $1$ .

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

One to any power is one.

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

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

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

Step 4

Verify if $f^{- 1} (x) = x^{3} + 3 x^{2} + 3 x + 1$ is the inverse of $f (x) = \sqrt[3]{x} - 1$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\sqrt[3]{x} - 1)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify each term.

Use the Binomial Theorem.

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

Simplify each term.

Rewrite ${\sqrt[3]{x}}^{3}$ as $x$ .

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

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

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

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

Combine $\frac{1}{3}$ and $3$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Rewrite ${\sqrt[3]{x}}^{2}$ as $\sqrt[3]{x^{2}}$ .

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

Multiply $- 1$ by $3$ .

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

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

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

Multiply $3$ by $1$ .

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

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

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

Rewrite ${(\sqrt[3]{x} - 1)}^{2}$ as $(\sqrt[3]{x} - 1) (\sqrt[3]{x} - 1)$ .

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

Expand $(\sqrt[3]{x} - 1) (\sqrt[3]{x} - 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt[3]{x} \sqrt[3]{x}$ .

Raise $\sqrt[3]{x}$ to the power of $1$ .

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

Raise $\sqrt[3]{x}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt[3]{x}}^{2}$ as $\sqrt[3]{x^{2}}$ .

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

Move $- 1$ to the left of $\sqrt[3]{x}$ .

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

Rewrite $- 1 \sqrt[3]{x}$ as $- \sqrt[3]{x}$ .

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

Rewrite $- 1 \sqrt[3]{x}$ as $- \sqrt[3]{x}$ .

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

Multiply $- 1$ by $- 1$ .

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

Subtract $\sqrt[3]{x}$ from $- \sqrt[3]{x}$ .

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

Apply the distributive property.

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

Simplify.

Multiply $- 2$ by $3$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x - 3 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} - 1 + 3 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} + 3 \cdot 1 + 3 (\sqrt[3]{x} - 1) + 1$

Multiply $3$ by $1$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x - 3 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} - 1 + 3 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} + 3 + 3 (\sqrt[3]{x} - 1) + 1$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} - 1) = x - 3 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} - 1 + 3 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} + 3 + 3 \sqrt[3]{x} + 3 \cdot - 1 + 1$

Multiply $3$ by $- 1$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x - 3 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} - 1 + 3 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} + 3 + 3 \sqrt[3]{x} - 3 + 1$

Simplify by adding terms.

Combine the opposite terms in $x - 3 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} - 1 + 3 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} + 3 + 3 \sqrt[3]{x} - 3 + 1$ .

Add $- 3 \sqrt[3]{x^{2}}$ and $3 \sqrt[3]{x^{2}}$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x + 0 + 3 \sqrt[3]{x} - 1 - 6 \sqrt[3]{x} + 3 + 3 \sqrt[3]{x} - 3 + 1$

Add $x$ and $0$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x + 3 \sqrt[3]{x} - 1 - 6 \sqrt[3]{x} + 3 + 3 \sqrt[3]{x} - 3 + 1$

Subtract $3$ from $3$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x + 3 \sqrt[3]{x} - 1 - 6 \sqrt[3]{x} + 0 + 3 \sqrt[3]{x} + 1$

Add $x + 3 \sqrt[3]{x} - 1 - 6 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x + 3 \sqrt[3]{x} - 1 - 6 \sqrt[3]{x} + 3 \sqrt[3]{x} + 1$

Add $- 1$ and $1$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x + 3 \sqrt[3]{x} + 0 - 6 \sqrt[3]{x} + 3 \sqrt[3]{x}$

Add $x + 3 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x + 3 \sqrt[3]{x} - 6 \sqrt[3]{x} + 3 \sqrt[3]{x}$

Subtract $6 \sqrt[3]{x}$ from $3 \sqrt[3]{x}$ .

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

Combine the opposite terms in $x - 3 \sqrt[3]{x} + 3 \sqrt[3]{x}$ .

Add $- 3 \sqrt[3]{x}$ and $3 \sqrt[3]{x}$ .

$f^{- 1} (\sqrt[3]{x} - 1) = x + 0$

Add $x$ and $0$ .

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (x^{3} + 3 x^{2} + 3 x + 1)$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Remove parentheses.

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

Simplify each term.

Factor $x^{3} + 3 x^{2} + 3 x + 1$ using the binomial theorem.

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

Pull terms out from under the radical, assuming real numbers.

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

Combine the opposite terms in $x + 1 - 1$ .

Subtract $1$ from $1$ .

$f (x^{3} + 3 x^{2} + 3 x + 1) = x + 0$

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{3} + 3 x^{2} + 3 x + 1$ is the inverse of $f (x) = \sqrt[3]{x} - 1$ .

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

Find the Inverse cube root of x-1

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