Find the Inverse f(x) = cube root of 8x+4

$f (x) = \sqrt[3]{8 x} + 4$

Step 1

Write $f (x) = \sqrt[3]{8 x} + 4$ as an equation.

$y = \sqrt[3]{8 x} + 4$

Step 2

Interchange the variables.

$x = \sqrt[3]{8 y} + 4$

Step 3

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{8 y} + 4 = x$ .

$\sqrt[3]{8 y} + 4 = x$

Subtract $4$ from both sides of the equation.

$\sqrt[3]{8 y} = x - 4$

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

${\sqrt[3]{8 y}}^{3} = {(x - 4)}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{8 y}$ as ${(8 y)}^{\frac{1}{3}}$ .

${({(8 y)}^{\frac{1}{3}})}^{3} = {(x - 4)}^{3}$

Simplify the left side.

Simplify ${({(8 y)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(8 y)}^{\frac{1}{3}})}^{3}$ .

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

${(8 y)}^{\frac{1}{3} \cdot 3} = {(x - 4)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(8 y)}^{\frac{1}{3} \cdot 3} = {(x - 4)}^{3}$

Rewrite the expression.

${(8 y)}^{1} = {(x - 4)}^{3}$

Simplify.

$8 y = {(x - 4)}^{3}$

Simplify the right side.

Simplify ${(x - 4)}^{3}$ .

Use the Binomial Theorem.

$8 y = x^{3} + 3 x^{2} \cdot - 4 + 3 x {(- 4)}^{2} + {(- 4)}^{3}$

Simplify each term.

Multiply $- 4$ by $3$ .

$8 y = x^{3} - 12 x^{2} + 3 x {(- 4)}^{2} + {(- 4)}^{3}$

Raise $- 4$ to the power of $2$ .

$8 y = x^{3} - 12 x^{2} + 3 x \cdot 16 + {(- 4)}^{3}$

Multiply $16$ by $3$ .

$8 y = x^{3} - 12 x^{2} + 48 x + {(- 4)}^{3}$

Raise $- 4$ to the power of $3$ .

$8 y = x^{3} - 12 x^{2} + 48 x - 64$

Divide each term in $8 y = x^{3} - 12 x^{2} + 48 x - 64$ by $8$ and simplify.

Divide each term in $8 y = x^{3} - 12 x^{2} + 48 x - 64$ by $8$ .

$\frac{8 y}{8} = \frac{x^{3}}{8} + \frac{- 12 x^{2}}{8} + \frac{48 x}{8} + \frac{- 64}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y}{8} = \frac{x^{3}}{8} + \frac{- 12 x^{2}}{8} + \frac{48 x}{8} + \frac{- 64}{8}$

Divide $y$ by $1$ .

$y = \frac{x^{3}}{8} + \frac{- 12 x^{2}}{8} + \frac{48 x}{8} + \frac{- 64}{8}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 12$ and $8$ .

Factor $4$ out of $- 12 x^{2}$ .

$y = \frac{x^{3}}{8} + \frac{4 (- 3 x^{2})}{8} + \frac{48 x}{8} + \frac{- 64}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

$y = \frac{x^{3}}{8} + \frac{4 (- 3 x^{2})}{4 (2)} + \frac{48 x}{8} + \frac{- 64}{8}$

Cancel the common factor.

$y = \frac{x^{3}}{8} + \frac{4 (- 3 x^{2})}{4 \cdot 2} + \frac{48 x}{8} + \frac{- 64}{8}$

Rewrite the expression.

$y = \frac{x^{3}}{8} + \frac{- 3 x^{2}}{2} + \frac{48 x}{8} + \frac{- 64}{8}$

Move the negative in front of the fraction.

$y = \frac{x^{3}}{8} - \frac{(3) x^{2}}{2} + \frac{48 x}{8} + \frac{- 64}{8}$

Cancel the common factor of $48$ and $8$ .

Factor $8$ out of $48 x$ .

$y = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + \frac{8 (6 x)}{8} + \frac{- 64}{8}$

Cancel the common factors.

Factor $8$ out of $8$ .

$y = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + \frac{8 (6 x)}{8 (1)} + \frac{- 64}{8}$

Cancel the common factor.

$y = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + \frac{8 (6 x)}{8 \cdot 1} + \frac{- 64}{8}$

Rewrite the expression.

$y = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + \frac{6 x}{1} + \frac{- 64}{8}$

Divide $6 x$ by $1$ .

$y = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x + \frac{- 64}{8}$

Divide $- 64$ by $8$ .

$y = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8$

Step 4

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

$f^{- 1} (x) = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8$

Step 5

Verify if $f^{- 1} (x) = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8$ is the inverse of $f (x) = \sqrt[3]{8 x} + 4$ .

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 $f^{- 1} (\sqrt[3]{8 x} + 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify each term.

Simplify the numerator.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{8 x}$ as ${(8 x)}^{\frac{1}{3}}$ .

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

Apply the product rule to $8 x$ .

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

Rewrite $8$ as $2^{3}$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Factor $2$ out of $2 x^{\frac{1}{3}} + 4$ .

Factor $2$ out of $2 x^{\frac{1}{3}}$ .

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

Factor $2$ out of $4$ .

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

Factor $2$ out of $2 x^{\frac{1}{3}} + 2 \cdot 2$ .

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

Apply the product rule to $2 (x^{\frac{1}{3}} + 2)$ .

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

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

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

Cancel the common factor.

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

Divide ${(x^{\frac{1}{3}} + 2)}^{3}$ by $1$ .

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

Use the Binomial Theorem.

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

Simplify each term.

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

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

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

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

Multiply $2$ by $3$ .

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

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

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

Multiply $4$ by $3$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 2^{3} - \frac{3 {(\sqrt[3]{8 x} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

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

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(\sqrt[3]{8 x} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Simplify the numerator.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{8 x}$ as ${(8 x)}^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {({(8 x)}^{\frac{1}{3}} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Apply the product rule to $8 x$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(8^{\frac{1}{3}} x^{\frac{1}{3}} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Rewrite $8$ as $2^{3}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {({(2^{3})}^{\frac{1}{3}} x^{\frac{1}{3}} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

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

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(2^{3 (\frac{1}{3})} x^{\frac{1}{3}} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(2^{3 (\frac{1}{3})} x^{\frac{1}{3}} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(2 x^{\frac{1}{3}} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Evaluate the exponent.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(2 x^{\frac{1}{3}} + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Factor $2$ out of $2 x^{\frac{1}{3}} + 4$ .

Factor $2$ out of $2 x^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(2 (x^{\frac{1}{3}}) + 4)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Factor $2$ out of $4$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(2 x^{\frac{1}{3}} + 2 \cdot 2)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Factor $2$ out of $2 x^{\frac{1}{3}} + 2 \cdot 2$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 {(2 (x^{\frac{1}{3}} + 2))}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Apply the product rule to $2 (x^{\frac{1}{3}} + 2)$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 \cdot (2^{2} {(x^{\frac{1}{3}} + 2)}^{2})}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

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

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{3 \cdot (4 {(x^{\frac{1}{3}} + 2)}^{2})}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Multiply $3$ by $4$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{12 {(x^{\frac{1}{3}} + 2)}^{2}}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Factor $2$ out of $12 {(x^{\frac{1}{3}} + 2)}^{2}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{2 (6 {(x^{\frac{1}{3}} + 2)}^{2})}{2} + 6 (\sqrt[3]{8 x} + 4) - 8$

Cancel the common factors.

Factor $2$ out of $2$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{2 (6 {(x^{\frac{1}{3}} + 2)}^{2})}{2 (1)} + 6 (\sqrt[3]{8 x} + 4) - 8$

Cancel the common factor.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{2 (6 {(x^{\frac{1}{3}} + 2)}^{2})}{2 \cdot 1} + 6 (\sqrt[3]{8 x} + 4) - 8$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - \frac{6 {(x^{\frac{1}{3}} + 2)}^{2}}{1} + 6 (\sqrt[3]{8 x} + 4) - 8$

Divide $6 {(x^{\frac{1}{3}} + 2)}^{2}$ by $1$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 {(x^{\frac{1}{3}} + 2)}^{2}) + 6 (\sqrt[3]{8 x} + 4) - 8$

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

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 ((x^{\frac{1}{3}} + 2) (x^{\frac{1}{3}} + 2))) + 6 (\sqrt[3]{8 x} + 4) - 8$

Expand $(x^{\frac{1}{3}} + 2) (x^{\frac{1}{3}} + 2)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{1}{3}} (x^{\frac{1}{3}} + 2) + 2 (x^{\frac{1}{3}} + 2))) + 6 (\sqrt[3]{8 x} + 4) - 8$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{1}{3}} x^{\frac{1}{3}} + x^{\frac{1}{3}} \cdot 2 + 2 (x^{\frac{1}{3}} + 2))) + 6 (\sqrt[3]{8 x} + 4) - 8$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{1}{3}} x^{\frac{1}{3}} + x^{\frac{1}{3}} \cdot 2 + 2 x^{\frac{1}{3}} + 2 \cdot 2)) + 6 (\sqrt[3]{8 x} + 4) - 8$

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.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{1}{3} + \frac{1}{3}} + x^{\frac{1}{3}} \cdot 2 + 2 x^{\frac{1}{3}} + 2 \cdot 2)) + 6 (\sqrt[3]{8 x} + 4) - 8$

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{1 + 1}{3}} + x^{\frac{1}{3}} \cdot 2 + 2 x^{\frac{1}{3}} + 2 \cdot 2)) + 6 (\sqrt[3]{8 x} + 4) - 8$

Add $1$ and $1$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{2}{3}} + x^{\frac{1}{3}} \cdot 2 + 2 x^{\frac{1}{3}} + 2 \cdot 2)) + 6 (\sqrt[3]{8 x} + 4) - 8$

Move $2$ to the left of $x^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{2}{3}} + 2 \cdot x^{\frac{1}{3}} + 2 x^{\frac{1}{3}} + 2 \cdot 2)) + 6 (\sqrt[3]{8 x} + 4) - 8$

Multiply $2$ by $2$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{2}{3}} + 2 x^{\frac{1}{3}} + 2 x^{\frac{1}{3}} + 4)) + 6 (\sqrt[3]{8 x} + 4) - 8$

Add $2 x^{\frac{1}{3}}$ and $2 x^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 (x^{\frac{2}{3}} + 4 x^{\frac{1}{3}} + 4)) + 6 (\sqrt[3]{8 x} + 4) - 8$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 x^{\frac{2}{3}} + 6 (4 x^{\frac{1}{3}}) + 6 \cdot 4) + 6 (\sqrt[3]{8 x} + 4) - 8$

Simplify.

Multiply $4$ by $6$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 x^{\frac{2}{3}} + 24 x^{\frac{1}{3}} + 6 \cdot 4) + 6 (\sqrt[3]{8 x} + 4) - 8$

Multiply $6$ by $4$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 x^{\frac{2}{3}} + 24 x^{\frac{1}{3}} + 24) + 6 (\sqrt[3]{8 x} + 4) - 8$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - (6 x^{\frac{2}{3}}) - (24 x^{\frac{1}{3}}) - 1 \cdot 24 + 6 (\sqrt[3]{8 x} + 4) - 8$

Simplify.

Multiply $6$ by $- 1$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - (24 x^{\frac{1}{3}}) - 1 \cdot 24 + 6 (\sqrt[3]{8 x} + 4) - 8$

Multiply $24$ by $- 1$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 1 \cdot 24 + 6 (\sqrt[3]{8 x} + 4) - 8$

Multiply $- 1$ by $24$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 24 + 6 (\sqrt[3]{8 x} + 4) - 8$

Simplify each term.

Rewrite $8$ as $2^{3}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 24 + 6 (\sqrt[3]{2^{3} x} + 4) - 8$

Pull terms out from under the radical.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 24 + 6 (2 \sqrt[3]{x} + 4) - 8$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 24 + 6 (2 \sqrt[3]{x}) + 6 \cdot 4 - 8$

Multiply $2$ by $6$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 24 + 12 \sqrt[3]{x} + 6 \cdot 4 - 8$

Multiply $6$ by $4$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 24 + 12 \sqrt[3]{x} + 24 - 8$

Simplify by adding terms.

Combine the opposite terms in $x + 6 x^{\frac{2}{3}} + 12 x^{\frac{1}{3}} + 8 - 6 x^{\frac{2}{3}} - 24 x^{\frac{1}{3}} - 24 + 12 \sqrt[3]{x} + 24 - 8$ .

Subtract $6 x^{\frac{2}{3}}$ from $6 x^{\frac{2}{3}}$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 12 x^{\frac{1}{3}} + 8 + 0 - 24 x^{\frac{1}{3}} - 24 + 12 \sqrt[3]{x} + 24 - 8$

Add $x + 12 x^{\frac{1}{3}} + 8$ and $0$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 12 x^{\frac{1}{3}} + 8 - 24 x^{\frac{1}{3}} - 24 + 12 \sqrt[3]{x} + 24 - 8$

Add $- 24$ and $24$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 12 x^{\frac{1}{3}} + 8 - 24 x^{\frac{1}{3}} + 0 + 12 \sqrt[3]{x} - 8$

Add $x + 12 x^{\frac{1}{3}} + 8 - 24 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 12 x^{\frac{1}{3}} + 8 - 24 x^{\frac{1}{3}} + 12 \sqrt[3]{x} - 8$

Subtract $8$ from $8$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 12 x^{\frac{1}{3}} + 0 - 24 x^{\frac{1}{3}} + 12 \sqrt[3]{x}$

Add $x + 12 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (\sqrt[3]{8 x} + 4) = x + 12 x^{\frac{1}{3}} - 24 x^{\frac{1}{3}} + 12 \sqrt[3]{x}$

Subtract $24 x^{\frac{1}{3}}$ from $12 x^{\frac{1}{3}}$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8)$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify each term.

To write $- \frac{3 x^{2}}{2}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3 x^{2}}{2}$ by $\frac{4}{4}$ .

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

Multiply $2$ by $4$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Factor $x^{2}$ out of $x^{3} - 3 x^{2} \cdot 4$ .

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

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

Factor $x^{2}$ out of $- 3 x^{2} \cdot 4$ .

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

Factor $x^{2}$ out of $x^{2} x + x^{2} (- 3 \cdot 4)$ .

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

Multiply $- 3$ by $4$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{2} (x - 12)}{8} + 6 x - 8)} + 4$

To write $6 x$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{2} (x - 12)}{8} + 6 x \cdot \frac{8}{8} - 8)} + 4$

Combine $6 x$ and $\frac{8}{8}$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{2} (x - 12)}{8} + \frac{6 x \cdot 8}{8} - 8)} + 4$

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{2} (x - 12) + 6 x \cdot 8}{8} - 8)} + 4$

Simplify the numerator.

Factor $x$ out of $x^{2} (x - 12) + 6 x \cdot 8$ .

Factor $x$ out of $x^{2} (x - 12)$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x (x - 12)) + 6 x \cdot 8}{8} - 8)} + 4$

Factor $x$ out of $6 x \cdot 8$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x (x - 12)) + x (6 \cdot 8)}{8} - 8)} + 4$

Factor $x$ out of $x (x (x - 12)) + x (6 \cdot 8)$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x (x - 12) + 6 \cdot 8)}{8} - 8)} + 4$

Apply the distributive property.

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x \cdot x + x \cdot - 12 + 6 \cdot 8)}{8} - 8)} + 4$

Multiply $x$ by $x$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x^{2} + x \cdot - 12 + 6 \cdot 8)}{8} - 8)} + 4$

Move $- 12$ to the left of $x$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x^{2} - 12 \cdot x + 6 \cdot 8)}{8} - 8)} + 4$

Multiply $6$ by $8$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x^{2} - 12 x + 48)}{8} - 8)} + 4$

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x^{2} - 12 x + 48)}{8} - 8 \cdot \frac{8}{8})} + 4$

Combine $- 8$ and $\frac{8}{8}$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x^{2} - 12 x + 48)}{8} + \frac{- 8 \cdot 8}{8})} + 4$

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x (x^{2} - 12 x + 48) - 8 \cdot 8}{8})} + 4$

Simplify the numerator.

Apply the distributive property.

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x \cdot x^{2} + x (- 12 x) + x \cdot 48 - 8 \cdot 8}{8})} + 4$

Simplify.

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

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

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

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x \cdot x^{2} + x (- 12 x) + x \cdot 48 - 8 \cdot 8}{8})} + 4$

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

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{1 + 2} + x (- 12 x) + x \cdot 48 - 8 \cdot 8}{8})} + 4$

Add $1$ and $2$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{3} + x (- 12 x) + x \cdot 48 - 8 \cdot 8}{8})} + 4$

Rewrite using the commutative property of multiplication.

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{3} - 12 x \cdot x + x \cdot 48 - 8 \cdot 8}{8})} + 4$

Move $48$ to the left of $x$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{3} - 12 x \cdot x + 48 \cdot x - 8 \cdot 8}{8})} + 4$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{3} - 12 (x \cdot x) + 48 \cdot x - 8 \cdot 8}{8})} + 4$

Multiply $x$ by $x$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{3} - 12 x^{2} + 48 \cdot x - 8 \cdot 8}{8})} + 4$

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{3} - 12 x^{2} + 48 x - 8 \cdot 8}{8})} + 4$

Multiply $- 8$ by $8$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{x^{3} - 12 x^{2} + 48 x - 64}{8})} + 4$

Rewrite $x^{3} - 12 x^{2} + 48 x - 64$ in a factored form.

Factor $x^{3} - 12 x^{2} + 48 x - 64$ 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.

$p = \pm 1, \pm 64, \pm 2, \pm 32, \pm 4, \pm 16, \pm 8 q = \pm 1$

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

$\pm 1, \pm 64, \pm 2, \pm 32, \pm 4, \pm 16, \pm 8$

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

Substitute $4$ into the polynomial.

$4^{3} - 12 \cdot 4^{2} + 48 \cdot 4 - 64$

Raise $4$ to the power of $3$ .

$64 - 12 \cdot 4^{2} + 48 \cdot 4 - 64$

Raise $4$ to the power of $2$ .

$64 - 12 \cdot 16 + 48 \cdot 4 - 64$

Multiply $- 12$ by $16$ .

$64 - 192 + 48 \cdot 4 - 64$

Subtract $192$ from $64$ .

$- 128 + 48 \cdot 4 - 64$

Multiply $48$ by $4$ .

$- 128 + 192 - 64$

Add $- 128$ and $192$ .

$64 - 64$

Subtract $64$ from $64$ .

$0$

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 12 x^{2} + 48 x - 64}{x - 4}$

Divide $x^{3} - 12 x^{2} + 48 x - 64$ by $x - 4$ .

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

$x$

$4$

$x^{3}$

$12 x^{2}$

$48 x$

$64$

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

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			+	$x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 4 x^{2}$

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$

Divide the highest order term in the dividend $- 8 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					-	$8 x^{2}$	+	$32 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 x^{2} + 32 x$

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$

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

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$

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

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$

Divide the highest order term in the dividend $16 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$
							+	$16 x$	-	$64$

The expression needs to be subtracted from the dividend, so change all the signs in $16 x - 64$

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$
							-	$16 x$	+	$64$

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

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$
							-	$16 x$	+	$64$
										$0$

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

$x^{2} - 8 x + 16$

Write $x^{3} - 12 x^{2} + 48 x - 64$ as a set of factors.

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = \sqrt[3]{8 (\frac{(x - 4) (x^{2} - 8 x + 16)}{8})} + 4$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

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

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

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

Combine like factors.

Raise $x - 4$ to the power of $1$ .

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

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

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

Add $1$ and $2$ .

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

Combine $8$ and $\frac{{(x - 4)}^{3}}{8}$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{8 {(x - 4)}^{3}}{8}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Divide ${(x - 4)}^{3}$ by $1$ .

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

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

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = x - 4 + 4$

Combine the opposite terms in $x - 4 + 4$ .

Add $- 4$ and $4$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = x + 0$

Add $x$ and $0$ .

$f (\frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8$ is the inverse of $f (x) = \sqrt[3]{8 x} + 4$ .

$f^{- 1} (x) = \frac{x^{3}}{8} - \frac{3 x^{2}}{2} + 6 x - 8$

Find the Inverse f(x) = cube root of 8x+4

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