Find the Inverse y = cube root of 3x+2+5

$y = \sqrt[3]{3 x + 2} + 5$

Interchange the variables.

$x = \sqrt[3]{3 y + 2} + 5$

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{3 y + 2} + 5 = x$ .

$\sqrt[3]{3 y + 2} + 5 = x$

Subtract $5$ from both sides of the equation.

$\sqrt[3]{3 y + 2} = x - 5$

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

${\sqrt[3]{3 y + 2}}^{3} = {(x - 5)}^{3}$

Simplify each side of the equation.

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

${({(3 y + 2)}^{\frac{1}{3}})}^{3} = {(x - 5)}^{3}$

Simplify the left side.

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

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

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

${(3 y + 2)}^{\frac{1}{3} \cdot 3} = {(x - 5)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(3 y + 2)}^{\frac{1}{3} \cdot 3} = {(x - 5)}^{3}$

Rewrite the expression.

${(3 y + 2)}^{1} = {(x - 5)}^{3}$

Simplify.

$3 y + 2 = {(x - 5)}^{3}$

Simplify the right side.

Simplify ${(x - 5)}^{3}$ .

Use the Binomial Theorem.

$3 y + 2 = x^{3} + 3 x^{2} \cdot - 5 + 3 x {(- 5)}^{2} + {(- 5)}^{3}$

Simplify each term.

Multiply $- 5$ by $3$ .

$3 y + 2 = x^{3} - 15 x^{2} + 3 x {(- 5)}^{2} + {(- 5)}^{3}$

Raise $- 5$ to the power of $2$ .

$3 y + 2 = x^{3} - 15 x^{2} + 3 x \cdot 25 + {(- 5)}^{3}$

Multiply $25$ by $3$ .

$3 y + 2 = x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}$

Raise $- 5$ to the power of $3$ .

$3 y + 2 = x^{3} - 15 x^{2} + 75 x - 125$

Solve for $y$ .

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

Subtract $2$ from both sides of the equation.

$3 y = x^{3} - 15 x^{2} + 75 x - 125 - 2$

Subtract $2$ from $- 125$ .

$3 y = x^{3} - 15 x^{2} + 75 x - 127$

Divide each term in $3 y = x^{3} - 15 x^{2} + 75 x - 127$ by $3$ and simplify.

Divide each term in $3 y = x^{3} - 15 x^{2} + 75 x - 127$ by $3$ .

$\frac{3 y}{3} = \frac{x^{3}}{3} + \frac{- 15 x^{2}}{3} + \frac{75 x}{3} + \frac{- 127}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{x^{3}}{3} + \frac{- 15 x^{2}}{3} + \frac{75 x}{3} + \frac{- 127}{3}$

Divide $y$ by $1$ .

$y = \frac{x^{3}}{3} + \frac{- 15 x^{2}}{3} + \frac{75 x}{3} + \frac{- 127}{3}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 15$ and $3$ .

Factor $3$ out of $- 15 x^{2}$ .

$y = \frac{x^{3}}{3} + \frac{3 (- 5 x^{2})}{3} + \frac{75 x}{3} + \frac{- 127}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$y = \frac{x^{3}}{3} + \frac{3 (- 5 x^{2})}{3 (1)} + \frac{75 x}{3} + \frac{- 127}{3}$

Cancel the common factor.

$y = \frac{x^{3}}{3} + \frac{3 (- 5 x^{2})}{3 \cdot 1} + \frac{75 x}{3} + \frac{- 127}{3}$

Rewrite the expression.

$y = \frac{x^{3}}{3} + \frac{- 5 x^{2}}{1} + \frac{75 x}{3} + \frac{- 127}{3}$

Divide $- 5 x^{2}$ by $1$ .

$y = \frac{x^{3}}{3} - 5 x^{2} + \frac{75 x}{3} + \frac{- 127}{3}$

Cancel the common factor of $75$ and $3$ .

Factor $3$ out of $75 x$ .

$y = \frac{x^{3}}{3} - 5 x^{2} + \frac{3 (25 x)}{3} + \frac{- 127}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$y = \frac{x^{3}}{3} - 5 x^{2} + \frac{3 (25 x)}{3 (1)} + \frac{- 127}{3}$

Cancel the common factor.

$y = \frac{x^{3}}{3} - 5 x^{2} + \frac{3 (25 x)}{3 \cdot 1} + \frac{- 127}{3}$

Rewrite the expression.

$y = \frac{x^{3}}{3} - 5 x^{2} + \frac{25 x}{1} + \frac{- 127}{3}$

Divide $25 x$ by $1$ .

$y = \frac{x^{3}}{3} - 5 x^{2} + 25 x + \frac{- 127}{3}$

Move the negative in front of the fraction.

$y = \frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}$

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

$f^{- 1} (x) = \frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}$

Verify if $f^{- 1} (x) = \frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}$ is the inverse of $f (x) = \sqrt[3]{3 x + 2} + 5$ .

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 $f^{- 1} (\sqrt[3]{3 x + 2} + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 {(\sqrt[3]{3 x + 2} + 5)}^{2} + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Simplify each term.

Rewrite ${(\sqrt[3]{3 x + 2} + 5)}^{2}$ as $(\sqrt[3]{3 x + 2} + 5) (\sqrt[3]{3 x + 2} + 5)$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 ((\sqrt[3]{3 x + 2} + 5) (\sqrt[3]{3 x + 2} + 5)) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Expand $(\sqrt[3]{3 x + 2} + 5) (\sqrt[3]{3 x + 2} + 5)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 (\sqrt[3]{3 x + 2} (\sqrt[3]{3 x + 2} + 5) + 5 (\sqrt[3]{3 x + 2} + 5)) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 (\sqrt[3]{3 x + 2} \sqrt[3]{3 x + 2} + \sqrt[3]{3 x + 2} \cdot 5 + 5 (\sqrt[3]{3 x + 2} + 5)) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 (\sqrt[3]{3 x + 2} \sqrt[3]{3 x + 2} + \sqrt[3]{3 x + 2} \cdot 5 + 5 \sqrt[3]{3 x + 2} + 5 \cdot 5) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Simplify and combine like terms.

Simplify each term.

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

Raise $\sqrt[3]{3 x + 2}$ to the power of $1$ .

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

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 ({\sqrt[3]{3 x + 2}}^{1 + 1} + \sqrt[3]{3 x + 2} \cdot 5 + 5 \sqrt[3]{3 x + 2} + 5 \cdot 5) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Add $1$ and $1$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 ({\sqrt[3]{3 x + 2}}^{2} + \sqrt[3]{3 x + 2} \cdot 5 + 5 \sqrt[3]{3 x + 2} + 5 \cdot 5) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

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

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 (\sqrt[3]{{(3 x + 2)}^{2}} + \sqrt[3]{3 x + 2} \cdot 5 + 5 \sqrt[3]{3 x + 2} + 5 \cdot 5) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Move $5$ to the left of $\sqrt[3]{3 x + 2}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 (\sqrt[3]{{(3 x + 2)}^{2}} + 5 \cdot \sqrt[3]{3 x + 2} + 5 \sqrt[3]{3 x + 2} + 5 \cdot 5) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Multiply $5$ by $5$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 (\sqrt[3]{{(3 x + 2)}^{2}} + 5 \sqrt[3]{3 x + 2} + 5 \sqrt[3]{3 x + 2} + 25) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Add $5 \sqrt[3]{3 x + 2}$ and $5 \sqrt[3]{3 x + 2}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 (\sqrt[3]{{(3 x + 2)}^{2}} + 10 \sqrt[3]{3 x + 2} + 25) + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 5 (10 \sqrt[3]{3 x + 2}) - 5 \cdot 25 + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Simplify.

Multiply $10$ by $- 5$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2} - 5 \cdot 25 + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Multiply $- 5$ by $25$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2} - 125 + 25 (\sqrt[3]{3 x + 2} + 5) - \frac{127}{3}$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2} - 125 + 25 \sqrt[3]{3 x + 2} + 25 \cdot 5 - \frac{127}{3}$

Multiply $25$ by $5$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2} - 125 + 25 \sqrt[3]{3 x + 2} + 125 - \frac{127}{3}$

Combine the opposite terms in $\frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2} - 125 + 25 \sqrt[3]{3 x + 2} + 125 - \frac{127}{3}$ .

Add $- 125$ and $125$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2} + 0 + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Add $\frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2}$ and $0$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

To write $- 5 \sqrt[3]{{(3 x + 2)}^{2}}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot \frac{3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Simplify terms.

Combine $- 5 \sqrt[3]{{(3 x + 2)}^{2}}$ and $\frac{3}{3}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3}}{3} + \frac{- 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(\sqrt[3]{3 x + 2} + 5)}^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Simplify the numerator.

Use the Binomial Theorem.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{\sqrt[3]{3 x + 2}}^{3} + 3 {\sqrt[3]{3 x + 2}}^{2} \cdot 5 + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Simplify each term.

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

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

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{({(3 x + 2)}^{\frac{1}{3}})}^{3} + 3 {\sqrt[3]{3 x + 2}}^{2} \cdot 5 + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

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

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(3 x + 2)}^{\frac{1}{3} \cdot 3} + 3 {\sqrt[3]{3 x + 2}}^{2} \cdot 5 + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

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

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{{(3 x + 2)}^{\frac{3}{3}} + 3 {\sqrt[3]{3 x + 2}}^{2} \cdot 5 + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

Rewrite the expression.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{(3 x + 2) + 3 {\sqrt[3]{3 x + 2}}^{2} \cdot 5 + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Simplify.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 2 + 3 {\sqrt[3]{3 x + 2}}^{2} \cdot 5 + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

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

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 2 + 3 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 5 + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Multiply $5$ by $3$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 2 + 15 \sqrt[3]{{(3 x + 2)}^{2}} + 3 \sqrt[3]{3 x + 2} \cdot 5^{2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Raise $5$ to the power of $2$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 2 + 15 \sqrt[3]{{(3 x + 2)}^{2}} + 3 \sqrt[3]{3 x + 2} \cdot 25 + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Multiply $25$ by $3$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 2 + 15 \sqrt[3]{{(3 x + 2)}^{2}} + 75 \sqrt[3]{3 x + 2} + 5^{3} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Raise $5$ to the power of $3$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 2 + 15 \sqrt[3]{{(3 x + 2)}^{2}} + 75 \sqrt[3]{3 x + 2} + 125 - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Add $2$ and $125$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 127 + 15 \sqrt[3]{{(3 x + 2)}^{2}} + 75 \sqrt[3]{3 x + 2} - 5 \sqrt[3]{{(3 x + 2)}^{2}} \cdot 3}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Multiply $3$ by $- 5$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 127 + 15 \sqrt[3]{{(3 x + 2)}^{2}} + 75 \sqrt[3]{3 x + 2} - 15 \sqrt[3]{{(3 x + 2)}^{2}}}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Subtract $15 \sqrt[3]{{(3 x + 2)}^{2}}$ from $15 \sqrt[3]{{(3 x + 2)}^{2}}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 127 + 0 + 75 \sqrt[3]{3 x + 2}}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Add $3 x$ and $0$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = \frac{3 x + 127 + 75 \sqrt[3]{3 x + 2}}{3} - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} - \frac{127}{3}$

Simplify terms.

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 x + 127 + 75 \sqrt[3]{3 x + 2} - 127}{3}$

Combine the opposite terms in $3 x + 127 + 75 \sqrt[3]{3 x + 2} - 127$ .

Subtract $127$ from $127$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 x + 0 + 75 \sqrt[3]{3 x + 2}}{3}$

Add $3 x$ and $0$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 x + 75 \sqrt[3]{3 x + 2}}{3}$

Cancel the common factor of $3 x + 75 \sqrt[3]{3 x + 2}$ and $3$ .

Factor $3$ out of $3 x$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 (x) + 75 \sqrt[3]{3 x + 2}}{3}$

Factor $3$ out of $75 \sqrt[3]{3 x + 2}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 (x) + 3 (25 \sqrt[3]{3 x + 2})}{3}$

Factor $3$ out of $3 (x) + 3 (25 \sqrt[3]{3 x + 2})$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 (x + 25 \sqrt[3]{3 x + 2})}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 (x + 25 \sqrt[3]{3 x + 2})}{3 (1)}$

Cancel the common factor.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{3 (x + 25 \sqrt[3]{3 x + 2})}{3 \cdot 1}$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + \frac{x + 25 \sqrt[3]{3 x + 2}}{1}$

Divide $x + 25 \sqrt[3]{3 x + 2}$ by $1$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 50 \sqrt[3]{3 x + 2} + 25 \sqrt[3]{3 x + 2} + x + 25 \sqrt[3]{3 x + 2}$

Add $- 50 \sqrt[3]{3 x + 2}$ and $25 \sqrt[3]{3 x + 2}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = - 25 \sqrt[3]{3 x + 2} + x + 25 \sqrt[3]{3 x + 2}$

Combine the opposite terms in $- 25 \sqrt[3]{3 x + 2} + x + 25 \sqrt[3]{3 x + 2}$ .

Add $- 25 \sqrt[3]{3 x + 2}$ and $25 \sqrt[3]{3 x + 2}$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\sqrt[3]{3 x + 2} + 5) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{3 (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) + 2} + 5$

Simplify each term.

Apply the distributive property.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{3 (\frac{x^{3}}{3}) + 3 (- 5 x^{2}) + 3 (25 x) + 3 (- \frac{127}{3}) + 2} + 5$

Simplify.

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{3 (\frac{x^{3}}{3}) + 3 (- 5 x^{2}) + 3 (25 x) + 3 (- \frac{127}{3}) + 2} + 5$

Rewrite the expression.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{x^{3} + 3 (- 5 x^{2}) + 3 (25 x) + 3 (- \frac{127}{3}) + 2} + 5$

Multiply $- 5$ by $3$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{x^{3} - 15 x^{2} + 3 (25 x) + 3 (- \frac{127}{3}) + 2} + 5$

Multiply $25$ by $3$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x + 3 (- \frac{127}{3}) + 2} + 5$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{127}{3}$ into the numerator.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x + 3 (\frac{- 127}{3}) + 2} + 5$

Cancel the common factor.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x + 3 (\frac{- 127}{3}) + 2} + 5$

Rewrite the expression.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x - 127 + 2} + 5$

Add $- 127$ and $2$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x - 125} + 5$

Rewrite $x^{3} - 15 x^{2} + 75 x - 125$ in a factored form.

Factor $x^{3} - 15 x^{2} + 75 x - 125$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 125, \pm 5, \pm 25 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 125, \pm 5, \pm 25$

Substitute $5$ and simplify the expression. In this case, the expression is equal to $0$ so $5$ is a root of the polynomial.

Substitute $5$ into the polynomial.

$5^{3} - 15 \cdot 5^{2} + 75 \cdot 5 - 125$

Raise $5$ to the power of $3$ .

$125 - 15 \cdot 5^{2} + 75 \cdot 5 - 125$

Raise $5$ to the power of $2$ .

$125 - 15 \cdot 25 + 75 \cdot 5 - 125$

Multiply $- 15$ by $25$ .

$125 - 375 + 75 \cdot 5 - 125$

Subtract $375$ from $125$ .

$- 250 + 75 \cdot 5 - 125$

Multiply $75$ by $5$ .

$- 250 + 375 - 125$

Add $- 250$ and $375$ .

$125 - 125$

Subtract $125$ from $125$ .

$0$

Since $5$ is a known root, divide the polynomial by $x - 5$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 15 x^{2} + 75 x - 125}{x - 5}$

Divide $x^{3} - 15 x^{2} + 75 x - 125$ by $x - 5$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$5$

$x^{3}$

$15 x^{2}$

$75 x$

$125$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			+	$x^{3}$	-	$5 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 5 x^{2}$

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$

Divide the highest order term in the dividend $- 10 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					-	$10 x^{2}$	+	$50 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 10 x^{2} + 50 x$

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$

Divide the highest order term in the dividend $25 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$
							+	$25 x$	-	$125$

The expression needs to be subtracted from the dividend, so change all the signs in $25 x - 125$

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$
							-	$25 x$	+	$125$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$
							-	$25 x$	+	$125$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - 10 x + 25$

Write $x^{3} - 15 x^{2} + 75 x - 125$ as a set of factors.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{(x - 5) (x^{2} - 10 x + 25)} + 5$

Factor using the perfect square rule.

Rewrite $25$ as $5^{2}$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{(x - 5) (x^{2} - 10 x + 5^{2})} + 5$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 x = 2 \cdot x \cdot 5$

Rewrite the polynomial.

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{(x - 5) (x^{2} - 2 \cdot x \cdot 5 + 5^{2})} + 5$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 5$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{(x - 5) {(x - 5)}^{2}} + 5$

Combine like factors.

Raise $x - 5$ to the power of $1$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{(x - 5) {(x - 5)}^{2}} + 5$

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

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{{(x - 5)}^{1 + 2}} + 5$

Add $1$ and $2$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = \sqrt[3]{{(x - 5)}^{3}} + 5$

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

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = x - 5 + 5$

Combine the opposite terms in $x - 5 + 5$ .

Add $- 5$ and $5$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = x + 0$

Add $x$ and $0$ .

$f (\frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}$ is the inverse of $f (x) = \sqrt[3]{3 x + 2} + 5$ .

$f^{- 1} (x) = \frac{x^{3}}{3} - 5 x^{2} + 25 x - \frac{127}{3}$

Find the Inverse y = cube root of 3x+2+5

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