Find the Inverse y = cube root of x-3

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Add $3$ to both sides of the equation.

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

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

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

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Use the Binomial Theorem.

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

Simplify each term.

Multiply $3$ by $3$ .

$y = x^{3} + 9 x^{2} + 3 x \cdot 3^{2} + 3^{3}$

Multiply $3$ by $3^{2}$ by adding the exponents.

Move $3^{2}$ .

$y = x^{3} + 9 x^{2} + 3^{2} \cdot 3 x + 3^{3}$

Multiply $3^{2}$ by $3$ .

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

$y = x^{3} + 9 x^{2} + 3^{2} \cdot 3^{1} x + 3^{3}$

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

$y = x^{3} + 9 x^{2} + 3^{2 + 1} x + 3^{3}$

Add $2$ and $1$ .

$y = x^{3} + 9 x^{2} + 3^{3} x + 3^{3}$

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

$y = x^{3} + 9 x^{2} + 27 x + 3^{3}$

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

$y = x^{3} + 9 x^{2} + 27 x + 27$

Step 3

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

$f^{- 1} (x) = x^{3} + 9 x^{2} + 27 x + 27$

Step 4

Verify if $f^{- 1} (x) = x^{3} + 9 x^{2} + 27 x + 27$ is the inverse of $f (x) = \sqrt[3]{x} - 3$ .

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

$f^{- 1} (\sqrt[3]{x} - 3) = {(\sqrt[3]{x} - 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

Simplify each term.

Use the Binomial Theorem.

$f^{- 1} (\sqrt[3]{x} - 3) = {\sqrt[3]{x}}^{3} + 3 {\sqrt[3]{x}}^{2} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

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 $x^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{x} - 3) = {(x^{\frac{1}{3}})}^{3} + 3 {\sqrt[3]{x}}^{2} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x^{\frac{1}{3} \cdot 3} + 3 {\sqrt[3]{x}}^{2} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x^{\frac{3}{3}} + 3 {\sqrt[3]{x}}^{2} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{x} - 3) = x^{\frac{3}{3}} + 3 {\sqrt[3]{x}}^{2} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{x} - 3) = x + 3 {\sqrt[3]{x}}^{2} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

Simplify.

$f^{- 1} (\sqrt[3]{x} - 3) = x + 3 {\sqrt[3]{x}}^{2} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x + 3 \sqrt[3]{x^{2}} \cdot - 3 + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

Multiply $- 3$ by $3$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

Raise $- 3$ to the power of $2$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} \cdot 9 + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

Multiply $9$ by $3$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} + {(- 3)}^{3} + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 {(\sqrt[3]{x} - 3)}^{2} + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 ((\sqrt[3]{x} - 3) (\sqrt[3]{x} - 3)) + 27 (\sqrt[3]{x} - 3) + 27$

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

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x} (\sqrt[3]{x} - 3) - 3 (\sqrt[3]{x} - 3)) + 27 (\sqrt[3]{x} - 3) + 27$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot - 3 - 3 (\sqrt[3]{x} - 3)) + 27 (\sqrt[3]{x} - 3) + 27$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot - 3 - 3 \sqrt[3]{x} - 3 \cdot - 3) + 27 (\sqrt[3]{x} - 3) + 27$

Simplify and combine like terms.

Simplify each term.

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

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot - 3 - 3 \sqrt[3]{x} - 3 \cdot - 3) + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot - 3 - 3 \sqrt[3]{x} - 3 \cdot - 3) + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 ({\sqrt[3]{x}}^{1 + 1} + \sqrt[3]{x} \cdot - 3 - 3 \sqrt[3]{x} - 3 \cdot - 3) + 27 (\sqrt[3]{x} - 3) + 27$

Add $1$ and $1$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 ({\sqrt[3]{x}}^{2} + \sqrt[3]{x} \cdot - 3 - 3 \sqrt[3]{x} - 3 \cdot - 3) + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot - 3 - 3 \sqrt[3]{x} - 3 \cdot - 3) + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x^{2}} - 3 \cdot \sqrt[3]{x} - 3 \sqrt[3]{x} - 3 \cdot - 3) + 27 (\sqrt[3]{x} - 3) + 27$

Multiply $- 3$ by $- 3$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x^{2}} - 3 \sqrt[3]{x} - 3 \sqrt[3]{x} + 9) + 27 (\sqrt[3]{x} - 3) + 27$

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

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 (\sqrt[3]{x^{2}} - 6 \sqrt[3]{x} + 9) + 27 (\sqrt[3]{x} - 3) + 27$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 \sqrt[3]{x^{2}} + 9 (- 6 \sqrt[3]{x}) + 9 \cdot 9 + 27 (\sqrt[3]{x} - 3) + 27$

Simplify.

Multiply $- 6$ by $9$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 \sqrt[3]{x^{2}} - 54 \sqrt[3]{x} + 9 \cdot 9 + 27 (\sqrt[3]{x} - 3) + 27$

Multiply $9$ by $9$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 \sqrt[3]{x^{2}} - 54 \sqrt[3]{x} + 81 + 27 (\sqrt[3]{x} - 3) + 27$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 \sqrt[3]{x^{2}} - 54 \sqrt[3]{x} + 81 + 27 \sqrt[3]{x} + 27 \cdot - 3 + 27$

Multiply $27$ by $- 3$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 \sqrt[3]{x^{2}} - 54 \sqrt[3]{x} + 81 + 27 \sqrt[3]{x} - 81 + 27$

Simplify by adding terms.

Combine the opposite terms in $x - 9 \sqrt[3]{x^{2}} + 27 \sqrt[3]{x} - 27 + 9 \sqrt[3]{x^{2}} - 54 \sqrt[3]{x} + 81 + 27 \sqrt[3]{x} - 81 + 27$ .

Add $- 9 \sqrt[3]{x^{2}}$ and $9 \sqrt[3]{x^{2}}$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x + 0 + 27 \sqrt[3]{x} - 27 - 54 \sqrt[3]{x} + 81 + 27 \sqrt[3]{x} - 81 + 27$

Add $x$ and $0$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x + 27 \sqrt[3]{x} - 27 - 54 \sqrt[3]{x} + 81 + 27 \sqrt[3]{x} - 81 + 27$

Subtract $81$ from $81$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x + 27 \sqrt[3]{x} - 27 - 54 \sqrt[3]{x} + 0 + 27 \sqrt[3]{x} + 27$

Add $x + 27 \sqrt[3]{x} - 27 - 54 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x + 27 \sqrt[3]{x} - 27 - 54 \sqrt[3]{x} + 27 \sqrt[3]{x} + 27$

Add $- 27$ and $27$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x + 27 \sqrt[3]{x} + 0 - 54 \sqrt[3]{x} + 27 \sqrt[3]{x}$

Add $x + 27 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x + 27 \sqrt[3]{x} - 54 \sqrt[3]{x} + 27 \sqrt[3]{x}$

Subtract $54 \sqrt[3]{x}$ from $27 \sqrt[3]{x}$ .

$f^{- 1} (\sqrt[3]{x} - 3) = x - 27 \sqrt[3]{x} + 27 \sqrt[3]{x}$

Combine the opposite terms in $x - 27 \sqrt[3]{x} + 27 \sqrt[3]{x}$ .

Add $- 27 \sqrt[3]{x}$ and $27 \sqrt[3]{x}$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (x^{3} + 9 x^{2} + 27 x + 27)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{3} + 9 x^{2} + 27 x + 27) = \sqrt[3]{x^{3} + 9 x^{2} + 27 x + 27} - 3$

Remove parentheses.

$f (x^{3} + 9 x^{2} + 27 x + 27) = \sqrt[3]{x^{3} + 9 x^{2} + 27 x + 27} - 3$

Simplify each term.

Make each term match the terms from the binomial theorem formula.

$f (x^{3} + 9 x^{2} + 27 x + 27) = \sqrt[3]{x^{3} + 3 \cdot (3 x^{2}) + 3 \cdot (3^{2} x) + 3^{3}} - 3$

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

$f (x^{3} + 9 x^{2} + 27 x + 27) = \sqrt[3]{{(x + 3)}^{3}} - 3$

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

$f (x^{3} + 9 x^{2} + 27 x + 27) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f (x^{3} + 9 x^{2} + 27 x + 27) = x + 0$

Add $x$ and $0$ .

$f (x^{3} + 9 x^{2} + 27 x + 27) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{3} + 9 x^{2} + 27 x + 27$ is the inverse of $f (x) = \sqrt[3]{x} - 3$ .

$f^{- 1} (x) = x^{3} + 9 x^{2} + 27 x + 27$

Find the Inverse y = cube root of x-3

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