Find the Inverse f(x)=(x^(1/3)-1)/9

$f (x) = \frac{x^{\frac{1}{3}} - 1}{9}$

Step 1

Write $f (x) = \frac{x^{\frac{1}{3}} - 1}{9}$ as an equation.

$y = \frac{x^{\frac{1}{3}} - 1}{9}$

Step 2

Interchange the variables.

$x = \frac{y^{\frac{1}{3}} - 1}{9}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{y^{\frac{1}{3}} - 1}{9} = x$ .

$\frac{y^{\frac{1}{3}} - 1}{9} = x$

Multiply both sides of the equation by $9$ .

$9 \frac{y^{\frac{1}{3}} - 1}{9} = 9 x$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$9 \frac{y^{\frac{1}{3}} - 1}{9} = 9 x$

Rewrite the expression.

$y^{\frac{1}{3}} - 1 = 9 x$

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

Add $1$ to both sides of the equation.

$y^{\frac{1}{3}} - 9 x = 1$

Add $9 x$ to both sides of the equation.

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

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

${(y^{\frac{1}{3}})}^{3} = {(1 + 9 x)}^{3}$

Simplify the exponent.

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

$y^{\frac{1}{3} \cdot 3} = {(1 + 9 x)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y^{\frac{1}{3} \cdot 3} = {(1 + 9 x)}^{3}$

Rewrite the expression.

$y^{1} = {(1 + 9 x)}^{3}$

Simplify.

$y = {(1 + 9 x)}^{3}$

Simplify the right side.

Simplify ${(1 + 9 x)}^{3}$ .

Use the Binomial Theorem.

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

Simplify each term.

One to any power is one.

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

One to any power is one.

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

Multiply $3$ by $1$ .

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

Multiply $9$ by $3$ .

$y = 1 + 27 x + 3 \cdot 1 {(9 x)}^{2} + {(9 x)}^{3}$

Multiply $3$ by $1$ .

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

Apply the product rule to $9 x$ .

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

Raise $9$ to the power of $2$ .

$y = 1 + 27 x + 3 (81 x^{2}) + {(9 x)}^{3}$

Multiply $81$ by $3$ .

$y = 1 + 27 x + 243 x^{2} + {(9 x)}^{3}$

Apply the product rule to $9 x$ .

$y = 1 + 27 x + 243 x^{2} + 9^{3} x^{3}$

Raise $9$ to the power of $3$ .

$y = 1 + 27 x + 243 x^{2} + 729 x^{3}$

Simplify $1 + 27 x + 243 x^{2} + 729 x^{3}$ .

Move $1$ .

$y = 27 x + 243 x^{2} + 729 x^{3} + 1$

Move $27 x$ .

$y = 243 x^{2} + 729 x^{3} + 27 x + 1$

Reorder $243 x^{2}$ and $729 x^{3}$ .

$y = 729 x^{3} + 243 x^{2} + 27 x + 1$

Step 4

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

$f^{- 1} (x) = 729 x^{3} + 243 x^{2} + 27 x + 1$

Step 5

Verify if $f^{- 1} (x) = 729 x^{3} + 243 x^{2} + 27 x + 1$ is the inverse of $f (x) = \frac{x^{\frac{1}{3}} - 1}{9}$ .

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 $f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = 729 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Simplify each term.

Apply the product rule to $\frac{x^{\frac{1}{3}} - 1}{9}$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = 729 (\frac{{(x^{\frac{1}{3}} - 1)}^{3}}{9^{3}}) + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Raise $9$ to the power of $3$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = 729 (\frac{{(x^{\frac{1}{3}} - 1)}^{3}}{729}) + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Cancel the common factor of $729$ .

Cancel the common factor.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = 729 (\frac{{(x^{\frac{1}{3}} - 1)}^{3}}{729}) + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Rewrite the expression.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = {(x^{\frac{1}{3}} - 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Use the Binomial Theorem.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = {(x^{\frac{1}{3}})}^{3} + 3 {(x^{\frac{1}{3}})}^{2} \cdot - 1 + 3 x^{\frac{1}{3}} {(- 1)}^{2} + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Simplify each term.

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

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x^{\frac{1}{3} \cdot 3} + 3 {(x^{\frac{1}{3}})}^{2} \cdot - 1 + 3 x^{\frac{1}{3}} {(- 1)}^{2} + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Cancel the common factor of $3$ .

Cancel the common factor.

Rewrite the expression.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 {(x^{\frac{1}{3}})}^{2} \cdot - 1 + 3 x^{\frac{1}{3}} {(- 1)}^{2} + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Simplify.

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

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3} \cdot 2} \cdot - 1 + 3 x^{\frac{1}{3}} {(- 1)}^{2} + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Combine $\frac{1}{3}$ and $2$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{2}{3}} \cdot - 1 + 3 x^{\frac{1}{3}} {(- 1)}^{2} + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Multiply $- 1$ by $3$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} {(- 1)}^{2} + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} \cdot 1 + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Multiply $3$ by $1$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} + {(- 1)}^{3} + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 243 {(\frac{x^{\frac{1}{3}} - 1}{9})}^{2} + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Apply the product rule to $\frac{x^{\frac{1}{3}} - 1}{9}$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 243 (\frac{{(x^{\frac{1}{3}} - 1)}^{2}}{9^{2}}) + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Raise $9$ to the power of $2$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 243 (\frac{{(x^{\frac{1}{3}} - 1)}^{2}}{81}) + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Cancel the common factor of $81$ .

Factor $81$ out of $243$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 81 (3) (\frac{{(x^{\frac{1}{3}} - 1)}^{2}}{81}) + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Cancel the common factor.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 81 \cdot (3 (\frac{{(x^{\frac{1}{3}} - 1)}^{2}}{81})) + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Rewrite the expression.

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

Rewrite ${(x^{\frac{1}{3}} - 1)}^{2}$ as $(x^{\frac{1}{3}} - 1) (x^{\frac{1}{3}} - 1)$ .

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

Expand $(x^{\frac{1}{3}} - 1) (x^{\frac{1}{3}} - 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x^{\frac{1}{3}}$ by $x^{\frac{1}{3}}$ by adding the exponents.

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $1$ .

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

Move $- 1$ to the left of $x^{\frac{1}{3}}$ .

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

Rewrite $- 1 x^{\frac{1}{3}}$ as $- x^{\frac{1}{3}}$ .

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

Rewrite $- 1 x^{\frac{1}{3}}$ as $- x^{\frac{1}{3}}$ .

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

Multiply $- 1$ by $- 1$ .

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

Subtract $x^{\frac{1}{3}}$ from $- x^{\frac{1}{3}}$ .

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

Apply the distributive property.

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

Simplify.

Multiply $- 2$ by $3$ .

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

Multiply $3$ by $1$ .

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

Cancel the common factor of $9$ .

Factor $9$ out of $27$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 9 (3) (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Cancel the common factor.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 9 \cdot (3 (\frac{x^{\frac{1}{3}} - 1}{9})) + 1$

Rewrite the expression.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 3 (x^{\frac{1}{3}} - 1) + 1$

Apply the distributive property.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} + 3 \cdot - 1 + 1$

Multiply $3$ by $- 1$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} - 3 + 1$

Simplify by adding terms.

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

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 + 0 - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} - 3 + 1$

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} - 3 + 1$

Subtract $3$ from $3$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}} + 0 + 1$

Add $x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}} + 1$

Add $- 1$ and $1$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}} + 0$

Add $x + 3 x^{\frac{1}{3}} - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}}$

Subtract $6 x^{\frac{1}{3}}$ from $3 x^{\frac{1}{3}}$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}}$

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

Add $- 3 x^{\frac{1}{3}}$ and $3 x^{\frac{1}{3}}$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x$

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

Set up the composite result function.

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

Evaluate $f (729 x^{3} + 243 x^{2} + 27 x + 1)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (729 x^{3} + 243 x^{2} + 27 x + 1) = \frac{{(729 x^{3} + 243 x^{2} + 27 x + 1)}^{\frac{1}{3}} - 1}{9}$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 729 x^{3} + 243 x^{2} + 27 x + 1$ is the inverse of $f (x) = \frac{x^{\frac{1}{3}} - 1}{9}$ .

$f^{- 1} (x) = 729 x^{3} + 243 x^{2} + 27 x + 1$

Find the Inverse f(x)=(x^(1/3)-1)/9

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