Find the Inverse 2 cube root of x-3

$2 \sqrt[3]{x} - 3$

Step 1

Interchange the variables.

$x = 2 \sqrt[3]{y} - 3$

Step 2

Solve for $y$ .

Rewrite the equation as $2 \sqrt[3]{y} - 3 = x$ .

$2 \sqrt[3]{y} - 3 = x$

Add $3$ to both sides of the equation.

$2 \sqrt[3]{y} = x + 3$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${(2 y^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $2 y^{\frac{1}{3}}$ .

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

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

$8 {(y^{\frac{1}{3}})}^{3} = {(x + 3)}^{3}$

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

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

$8 y^{\frac{1}{3} \cdot 3} = {(x + 3)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 y^{\frac{1}{3} \cdot 3} = {(x + 3)}^{3}$

Rewrite the expression.

$8 y^{1} = {(x + 3)}^{3}$

Simplify.

$8 y = {(x + 3)}^{3}$

Simplify the right side.

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

Use the Binomial Theorem.

$8 y = x^{3} + 3 x^{2} \cdot 3 + 3 x \cdot 3^{2} + 3^{3}$

Simplify each term.

Multiply $3$ by $3$ .

$8 y = x^{3} + 9 x^{2} + 3 x \cdot 3^{2} + 3^{3}$

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

Move $3^{2}$ .

$8 y = x^{3} + 9 x^{2} + 3^{2} \cdot 3 x + 3^{3}$

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

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

$8 y = x^{3} + 9 x^{2} + 3^{2} \cdot 3^{1} x + 3^{3}$

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

$8 y = x^{3} + 9 x^{2} + 3^{2 + 1} x + 3^{3}$

Add $2$ and $1$ .

$8 y = x^{3} + 9 x^{2} + 3^{3} x + 3^{3}$

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

$8 y = x^{3} + 9 x^{2} + 27 x + 3^{3}$

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

$8 y = x^{3} + 9 x^{2} + 27 x + 27$

Divide each term in $8 y = x^{3} + 9 x^{2} + 27 x + 27$ by $8$ and simplify.

Divide each term in $8 y = x^{3} + 9 x^{2} + 27 x + 27$ by $8$ .

$\frac{8 y}{8} = \frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y}{8} = \frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}$

Divide $y$ by $1$ .

$y = \frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}$

Step 3

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

$f^{- 1} (x) = \frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}$ is the inverse of $f (x) = 2 \sqrt[3]{x} - 3$ .

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 $f^{- 1} (2 \sqrt[3]{x} - 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{{(2 \sqrt[3]{x} - 3)}^{3}}{8} + \frac{9 {(2 \sqrt[3]{x} - 3)}^{2}}{8} + \frac{27 (2 \sqrt[3]{x} - 3)}{8} + \frac{27}{8}$

Combine the numerators over the common denominator.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{{(2 \sqrt[3]{x} - 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Simplify each term.

Use the Binomial Theorem.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{{(2 \sqrt[3]{x})}^{3} + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Simplify each term.

Apply the product rule to $2 \sqrt[3]{x}$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{2^{3} {\sqrt[3]{x}}^{3} + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 {\sqrt[3]{x}}^{3} + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 {(x^{\frac{1}{3}})}^{3} + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x^{\frac{1}{3} \cdot 3} + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x^{\frac{3}{3}} + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Cancel the common factor of $3$ .

Cancel the common factor.

Rewrite the expression.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Simplify.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 3 {(2 \sqrt[3]{x})}^{2} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Apply the product rule to $2 \sqrt[3]{x}$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 3 (2^{2} {\sqrt[3]{x}}^{2}) \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 3 (4 {\sqrt[3]{x}}^{2}) \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 3 (4 \sqrt[3]{x^{2}}) \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $4$ by $3$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 12 \sqrt[3]{x^{2}} \cdot - 3 + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $- 3$ by $12$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 3 (2 \sqrt[3]{x}) {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $2$ by $3$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 6 \sqrt[3]{x} {(- 3)}^{2} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 6 \sqrt[3]{x} \cdot 9 + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $9$ by $6$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} + {(- 3)}^{3} + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 {(2 \sqrt[3]{x} - 3)}^{2} + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 ((2 \sqrt[3]{x} - 3) (2 \sqrt[3]{x} - 3)) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Expand $(2 \sqrt[3]{x} - 3) (2 \sqrt[3]{x} - 3)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (2 \sqrt[3]{x} (2 \sqrt[3]{x} - 3) - 3 (2 \sqrt[3]{x} - 3)) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Apply the distributive property.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (2 \sqrt[3]{x} (2 \sqrt[3]{x}) + 2 \sqrt[3]{x} \cdot - 3 - 3 (2 \sqrt[3]{x} - 3)) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Apply the distributive property.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (2 \sqrt[3]{x} (2 \sqrt[3]{x}) + 2 \sqrt[3]{x} \cdot - 3 - 3 (2 \sqrt[3]{x}) - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Simplify and combine like terms.

Simplify each term.

Multiply $2 \sqrt[3]{x} (2 \sqrt[3]{x})$ .

Multiply $2$ by $2$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 \sqrt[3]{x} \sqrt[3]{x} + 2 \sqrt[3]{x} \cdot - 3 - 3 (2 \sqrt[3]{x}) - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 (\sqrt[3]{x} \sqrt[3]{x}) + 2 \sqrt[3]{x} \cdot - 3 - 3 (2 \sqrt[3]{x}) - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 {\sqrt[3]{x}}^{1 + 1} + 2 \sqrt[3]{x} \cdot - 3 - 3 (2 \sqrt[3]{x}) - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Add $1$ and $1$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 {\sqrt[3]{x}}^{2} + 2 \sqrt[3]{x} \cdot - 3 - 3 (2 \sqrt[3]{x}) - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 \sqrt[3]{x^{2}} + 2 \sqrt[3]{x} \cdot - 3 - 3 (2 \sqrt[3]{x}) - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $- 3$ by $2$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} - 3 (2 \sqrt[3]{x}) - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $2$ by $- 3$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} - 6 \sqrt[3]{x} - 3 \cdot - 3) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $- 3$ by $- 3$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 \sqrt[3]{x^{2}} - 6 \sqrt[3]{x} - 6 \sqrt[3]{x} + 9) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 \sqrt[3]{x^{2}} - 12 \sqrt[3]{x} + 9) + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Apply the distributive property.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 9 (4 \sqrt[3]{x^{2}}) + 9 (- 12 \sqrt[3]{x}) + 9 \cdot 9 + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Simplify.

Multiply $4$ by $9$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 36 \sqrt[3]{x^{2}} + 9 (- 12 \sqrt[3]{x}) + 9 \cdot 9 + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $- 12$ by $9$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 36 \sqrt[3]{x^{2}} - 108 \sqrt[3]{x} + 9 \cdot 9 + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Multiply $9$ by $9$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 36 \sqrt[3]{x^{2}} - 108 \sqrt[3]{x} + 81 + 27 (2 \sqrt[3]{x} - 3) + 27}{8}$

Apply the distributive property.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 36 \sqrt[3]{x^{2}} - 108 \sqrt[3]{x} + 81 + 27 (2 \sqrt[3]{x}) + 27 \cdot - 3 + 27}{8}$

Multiply $2$ by $27$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 36 \sqrt[3]{x^{2}} - 108 \sqrt[3]{x} + 81 + 54 \sqrt[3]{x} + 27 \cdot - 3 + 27}{8}$

Multiply $27$ by $- 3$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 36 \sqrt[3]{x^{2}} - 108 \sqrt[3]{x} + 81 + 54 \sqrt[3]{x} - 81 + 27}{8}$

Simplify terms.

Combine the opposite terms in $8 x - 36 \sqrt[3]{x^{2}} + 54 \sqrt[3]{x} - 27 + 36 \sqrt[3]{x^{2}} - 108 \sqrt[3]{x} + 81 + 54 \sqrt[3]{x} - 81 + 27$ .

Add $- 36 \sqrt[3]{x^{2}}$ and $36 \sqrt[3]{x^{2}}$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 0 + 54 \sqrt[3]{x} - 27 - 108 \sqrt[3]{x} + 81 + 54 \sqrt[3]{x} - 81 + 27}{8}$

Add $8 x$ and $0$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 54 \sqrt[3]{x} - 27 - 108 \sqrt[3]{x} + 81 + 54 \sqrt[3]{x} - 81 + 27}{8}$

Subtract $81$ from $81$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 54 \sqrt[3]{x} - 27 - 108 \sqrt[3]{x} + 0 + 54 \sqrt[3]{x} + 27}{8}$

Add $8 x + 54 \sqrt[3]{x} - 27 - 108 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 54 \sqrt[3]{x} - 27 - 108 \sqrt[3]{x} + 54 \sqrt[3]{x} + 27}{8}$

Add $- 27$ and $27$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 54 \sqrt[3]{x} + 0 - 108 \sqrt[3]{x} + 54 \sqrt[3]{x}}{8}$

Add $8 x + 54 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 54 \sqrt[3]{x} - 108 \sqrt[3]{x} + 54 \sqrt[3]{x}}{8}$

Subtract $108 \sqrt[3]{x}$ from $54 \sqrt[3]{x}$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x - 54 \sqrt[3]{x} + 54 \sqrt[3]{x}}{8}$

Combine the opposite terms in $8 x - 54 \sqrt[3]{x} + 54 \sqrt[3]{x}$ .

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

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x + 0}{8}$

Add $8 x$ and $0$ .

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x}{8}$

Cancel the common factor of $8$ .

Cancel the common factor.

$f^{- 1} (2 \sqrt[3]{x} - 3) = \frac{8 x}{8}$

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 \sqrt[3]{\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}} - 3$

Remove parentheses.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 \sqrt[3]{\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}} - 3$

Simplify each term.

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 \sqrt[3]{\frac{x^{3} + 9 x^{2} + 27 x + 27}{8}} - 3$

Rewrite $\frac{x^{3} + 9 x^{2} + 27 x + 27}{8}$ as ${(\frac{1}{2})}^{3} (x^{3} + 9 x^{2} + 27 x + 27)$ .

Factor the perfect power $1^{3}$ out of $x^{3} + 9 x^{2} + 27 x + 27$ .

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 \sqrt[3]{\frac{1^{3} (x^{3} + 9 x^{2} + 27 x + 27)}{8}} - 3$

Factor the perfect power $2^{3}$ out of $8$ .

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 \sqrt[3]{\frac{1^{3} (x^{3} + 9 x^{2} + 27 x + 27)}{2^{3} \cdot 1}} - 3$

Rearrange the fraction $\frac{1^{3} (x^{3} + 9 x^{2} + 27 x + 27)}{2^{3} \cdot 1}$ .

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 \sqrt[3]{{(\frac{1}{2})}^{3} (x^{3} + 9 x^{2} + 27 x + 27)} - 3$

Pull terms out from under the radical.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{1}{2} \cdot \sqrt[3]{x^{3} + 9 x^{2} + 27 x + 27}) - 3$

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

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{1}{2} \cdot \sqrt[3]{x^{3} + 3 \cdot (3 x^{2}) + 3 \cdot (3^{2} x) + 3^{3}}) - 3$

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

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{1}{2} \cdot \sqrt[3]{{(x + 3)}^{3}}) - 3$

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

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{1}{2} \cdot (x + 3)) - 3$

Apply the distributive property.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{1}{2} x + \frac{1}{2} \cdot 3) - 3$

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

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{x}{2} + \frac{1}{2} \cdot 3) - 3$

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

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{x}{2} + \frac{3}{2}) - 3$

Apply the distributive property.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{x}{2}) + 2 (\frac{3}{2}) - 3$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = 2 (\frac{x}{2}) + 2 (\frac{3}{2}) - 3$

Rewrite the expression.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = x + 2 (\frac{3}{2}) - 3$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = x + 2 (\frac{3}{2}) - 3$

Rewrite the expression.

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = x + 3 - 3$

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

Subtract $3$ from $3$ .

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = x + 0$

Add $x$ and $0$ .

$f (\frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}$ is the inverse of $f (x) = 2 \sqrt[3]{x} - 3$ .

$f^{- 1} (x) = \frac{x^{3}}{8} + \frac{9 x^{2}}{8} + \frac{27 x}{8} + \frac{27}{8}$

Find the Inverse 2 cube root of x-3

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