Precalculus Examples

$f (x) = 8 \sqrt[3]{x} - 4$ ; find $f^{- 1} (x)$

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Add $4$ to both sides of the equation.

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

Step 3.3

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

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

Step 3.4

Simplify each side of the equation.

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Step 3.4.1

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

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

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

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

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Step 3.4.2.1.1

Apply the product rule to $8 y^{\frac{1}{3}}$ .

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

Step 3.4.2.1.2

Raise $8$ to the power of $3$ .

$512 {(y^{\frac{1}{3}})}^{3} = {(x + 4)}^{3}$

Step 3.4.2.1.3

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

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Step 3.4.2.1.3.1

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

$512 y^{\frac{1}{3} \cdot 3} = {(x + 4)}^{3}$

Step 3.4.2.1.3.2

Cancel the common factor of $3$ .

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Step 3.4.2.1.3.2.1

Cancel the common factor.

$512 y^{\frac{1}{3} \cdot 3} = {(x + 4)}^{3}$

Step 3.4.2.1.3.2.2

Rewrite the expression.

$512 y^{1} = {(x + 4)}^{3}$

Step 3.4.2.1.4

Simplify.

$512 y = {(x + 4)}^{3}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

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

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Step 3.4.3.1.1

Use the Binomial Theorem.

$512 y = x^{3} + 3 x^{2} \cdot 4 + 3 x \cdot 4^{2} + 4^{3}$

Step 3.4.3.1.2

Simplify each term.

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Step 3.4.3.1.2.1

Multiply $4$ by $3$ .

$512 y = x^{3} + 12 x^{2} + 3 x \cdot 4^{2} + 4^{3}$

Step 3.4.3.1.2.2

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

$512 y = x^{3} + 12 x^{2} + 3 x \cdot 16 + 4^{3}$

Step 3.4.3.1.2.3

Multiply $16$ by $3$ .

$512 y = x^{3} + 12 x^{2} + 48 x + 4^{3}$

Step 3.4.3.1.2.4

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

$512 y = x^{3} + 12 x^{2} + 48 x + 64$

Step 3.5

Divide each term in $512 y = x^{3} + 12 x^{2} + 48 x + 64$ by $512$ and simplify.

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Step 3.5.1

Divide each term in $512 y = x^{3} + 12 x^{2} + 48 x + 64$ by $512$ .

$\frac{512 y}{512} = \frac{x^{3}}{512} + \frac{12 x^{2}}{512} + \frac{48 x}{512} + \frac{64}{512}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $512$ .

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Step 3.5.2.1.1

Cancel the common factor.

$\frac{512 y}{512} = \frac{x^{3}}{512} + \frac{12 x^{2}}{512} + \frac{48 x}{512} + \frac{64}{512}$

Step 3.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{x^{3}}{512} + \frac{12 x^{2}}{512} + \frac{48 x}{512} + \frac{64}{512}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Simplify each term.

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Step 3.5.3.1.1

Cancel the common factor of $12$ and $512$ .

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Step 3.5.3.1.1.1

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

$y = \frac{x^{3}}{512} + \frac{4 (3 x^{2})}{512} + \frac{48 x}{512} + \frac{64}{512}$

Step 3.5.3.1.1.2

Cancel the common factors.

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Step 3.5.3.1.1.2.1

Factor $4$ out of $512$ .

$y = \frac{x^{3}}{512} + \frac{4 (3 x^{2})}{4 (128)} + \frac{48 x}{512} + \frac{64}{512}$

Step 3.5.3.1.1.2.2

Cancel the common factor.

$y = \frac{x^{3}}{512} + \frac{4 (3 x^{2})}{4 \cdot 128} + \frac{48 x}{512} + \frac{64}{512}$

Step 3.5.3.1.1.2.3

Rewrite the expression.

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{48 x}{512} + \frac{64}{512}$

Step 3.5.3.1.2

Cancel the common factor of $48$ and $512$ .

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Step 3.5.3.1.2.1

Factor $16$ out of $48 x$ .

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{16 (3 x)}{512} + \frac{64}{512}$

Step 3.5.3.1.2.2

Cancel the common factors.

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Step 3.5.3.1.2.2.1

Factor $16$ out of $512$ .

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{16 (3 x)}{16 (32)} + \frac{64}{512}$

Step 3.5.3.1.2.2.2

Cancel the common factor.

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{16 (3 x)}{16 \cdot 32} + \frac{64}{512}$

Step 3.5.3.1.2.2.3

Rewrite the expression.

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{64}{512}$

Step 3.5.3.1.3

Cancel the common factor of $64$ and $512$ .

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Step 3.5.3.1.3.1

Factor $64$ out of $64$ .

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{64 (1)}{512}$

Step 3.5.3.1.3.2

Cancel the common factors.

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Step 3.5.3.1.3.2.1

Factor $64$ out of $512$ .

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{64 \cdot 1}{64 \cdot 8}$

Step 3.5.3.1.3.2.2

Cancel the common factor.

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{64 \cdot 1}{64 \cdot 8}$

Step 3.5.3.1.3.2.3

Rewrite the expression.

$y = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}$

Step 4

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

$f^{- 1} (x) = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}$

Step 5

Verify if $f^{- 1} (x) = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}$ is the inverse of $f (x) = 8 \sqrt[3]{x} - 4$ .

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Step 5.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(8 \sqrt[3]{x} - 4)}^{3}}{512} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3

Simplify terms.

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Step 5.2.3.1

Simplify each term.

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Step 5.2.3.1.1

Simplify the numerator.

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Step 5.2.3.1.1.1

Factor $4$ out of $8 \sqrt[3]{x} - 4$ .

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Step 5.2.3.1.1.1.1

Factor $4$ out of $8 \sqrt[3]{x}$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(4 (2 \sqrt[3]{x}) - 4)}^{3}}{512} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.1.1.2

Factor $4$ out of $- 4$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(4 (2 \sqrt[3]{x}) + 4 (- 1))}^{3}}{512} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.1.1.3

Factor $4$ out of $4 (2 \sqrt[3]{x}) + 4 (- 1)$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(4 (2 \sqrt[3]{x} - 1))}^{3}}{512} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.1.2

Apply the product rule to $4 (2 \sqrt[3]{x} - 1)$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{4^{3} {(2 \sqrt[3]{x} - 1)}^{3}}{512} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.1.3

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

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{64 {(2 \sqrt[3]{x} - 1)}^{3}}{512} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.2

Cancel the common factor of $64$ and $512$ .

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Step 5.2.3.1.2.1

Factor $64$ out of $64 {(2 \sqrt[3]{x} - 1)}^{3}$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{64 ({(2 \sqrt[3]{x} - 1)}^{3})}{512} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.2.2

Cancel the common factors.

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Step 5.2.3.1.2.2.1

Factor $64$ out of $512$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{64 {(2 \sqrt[3]{x} - 1)}^{3}}{64 \cdot 8} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.2.2.2

Cancel the common factor.

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{64 {(2 \sqrt[3]{x} - 1)}^{3}}{64 \cdot 8} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.2.2.3

Rewrite the expression.

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 {(8 \sqrt[3]{x} - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.3

Simplify the numerator.

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Step 5.2.3.1.3.1

Factor $4$ out of $8 \sqrt[3]{x} - 4$ .

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Step 5.2.3.1.3.1.1

Factor $4$ out of $8 \sqrt[3]{x}$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 {(4 (2 \sqrt[3]{x}) - 4)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.3.1.2

Factor $4$ out of $- 4$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 {(4 (2 \sqrt[3]{x}) + 4 (- 1))}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.3.1.3

Factor $4$ out of $4 (2 \sqrt[3]{x}) + 4 (- 1)$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 {(4 (2 \sqrt[3]{x} - 1))}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.3.2

Apply the product rule to $4 (2 \sqrt[3]{x} - 1)$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 \cdot (4^{2} {(2 \sqrt[3]{x} - 1)}^{2})}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.3.3

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

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 \cdot (16 {(2 \sqrt[3]{x} - 1)}^{2})}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.3.4

Multiply $3$ by $16$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{48 {(2 \sqrt[3]{x} - 1)}^{2}}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.4

Cancel the common factor of $48$ and $128$ .

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Step 5.2.3.1.4.1

Factor $16$ out of $48 {(2 \sqrt[3]{x} - 1)}^{2}$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{16 (3 {(2 \sqrt[3]{x} - 1)}^{2})}{128} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.4.2

Cancel the common factors.

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Step 5.2.3.1.4.2.1

Factor $16$ out of $128$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{16 (3 {(2 \sqrt[3]{x} - 1)}^{2})}{16 (8)} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.4.2.2

Cancel the common factor.

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{16 (3 {(2 \sqrt[3]{x} - 1)}^{2})}{16 \cdot 8} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.4.2.3

Rewrite the expression.

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 {(2 \sqrt[3]{x} - 1)}^{2}}{8} + \frac{3 (8 \sqrt[3]{x} - 4)}{32} + \frac{1}{8}$

Step 5.2.3.1.5

Cancel the common factor of $8 \sqrt[3]{x} - 4$ and $32$ .

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Step 5.2.3.1.5.1

Factor $4$ out of $3 (8 \sqrt[3]{x} - 4)$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3}}{8} + \frac{3 {(2 \sqrt[3]{x} - 1)}^{2}}{8} + \frac{4 (3 (2 \sqrt[3]{x} - 1))}{32} + \frac{1}{8}$

Step 5.2.3.1.5.2

Cancel the common factors.

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Step 5.2.3.1.5.2.1

Factor $4$ out of $32$ .

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

Step 5.2.3.1.5.2.2

Cancel the common factor.

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

Step 5.2.3.1.5.2.3

Rewrite the expression.

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

Step 5.2.3.2

Combine the numerators over the common denominator.

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

Step 5.2.3.3

Simplify each term.

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Step 5.2.3.3.1

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

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

Step 5.2.3.3.2

Expand $(2 \sqrt[3]{x} - 1) (2 \sqrt[3]{x} - 1)$ using the FOIL Method.

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Step 5.2.3.3.2.1

Apply the distributive property.

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

Step 5.2.3.3.2.2

Apply the distributive property.

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

Step 5.2.3.3.2.3

Apply the distributive property.

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

Step 5.2.3.3.3

Simplify and combine like terms.

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Step 5.2.3.3.3.1

Simplify each term.

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Step 5.2.3.3.3.1.1

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

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Step 5.2.3.3.3.1.1.1

Multiply $2$ by $2$ .

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

Step 5.2.3.3.3.1.1.2

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

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

Step 5.2.3.3.3.1.1.3

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

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

Step 5.2.3.3.3.1.1.4

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

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

Step 5.2.3.3.3.1.1.5

Add $1$ and $1$ .

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

Step 5.2.3.3.3.1.2

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

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

Step 5.2.3.3.3.1.3

Multiply $- 1$ by $2$ .

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

Step 5.2.3.3.3.1.4

Multiply $2$ by $- 1$ .

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

Step 5.2.3.3.3.1.5

Multiply $- 1$ by $- 1$ .

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

Step 5.2.3.3.3.2

Subtract $2 \sqrt[3]{x}$ from $- 2 \sqrt[3]{x}$ .

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

Step 5.2.3.3.4

Apply the distributive property.

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

Step 5.2.3.3.5

Simplify.

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Step 5.2.3.3.5.1

Multiply $4$ by $3$ .

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

Step 5.2.3.3.5.2

Multiply $- 4$ by $3$ .

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

Step 5.2.3.3.5.3

Multiply $3$ by $1$ .

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

Step 5.2.3.3.6

Apply the distributive property.

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

Step 5.2.3.3.7

Multiply $2$ by $3$ .

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

Step 5.2.3.3.8

Multiply $3$ by $- 1$ .

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

Step 5.2.3.4

Simplify by adding terms.

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Step 5.2.3.4.1

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

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Step 5.2.3.4.1.1

Subtract $3$ from $3$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{{(2 \sqrt[3]{x} - 1)}^{3} + 12 \sqrt[3]{x^{2}} - 12 \sqrt[3]{x} + 0 + 6 \sqrt[3]{x} + 1}{8}$

Step 5.2.3.4.1.2

Add ${(2 \sqrt[3]{x} - 1)}^{3} + 12 \sqrt[3]{x^{2}} - 12 \sqrt[3]{x}$ and $0$ .

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

Step 5.2.3.4.2

Add $- 12 \sqrt[3]{x}$ and $6 \sqrt[3]{x}$ .

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

Step 5.2.4

Simplify the numerator.

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Step 5.2.4.1

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

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

Step 5.2.4.2

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

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

Step 5.2.4.3

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

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

Step 5.2.4.4

Use the Binomial Theorem.

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

Step 5.2.4.5

Simplify each term.

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Step 5.2.4.5.1

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

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

Step 5.2.4.5.2

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

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

Step 5.2.4.5.3

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

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Step 5.2.4.5.3.1

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

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

Step 5.2.4.5.3.2

Cancel the common factor of $3$ .

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Step 5.2.4.5.3.2.1

Cancel the common factor.

Step 5.2.4.5.3.2.2

Rewrite the expression.

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

Step 5.2.4.5.4

Simplify.

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

Step 5.2.4.5.5

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

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

Step 5.2.4.5.6

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

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

Step 5.2.4.5.7

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

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Step 5.2.4.5.7.1

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

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

Step 5.2.4.5.7.2

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

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

Step 5.2.4.5.8

Multiply $4$ by $3$ .

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

Step 5.2.4.5.9

Multiply $- 1$ by $12$ .

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

Step 5.2.4.5.10

Multiply $2$ by $3$ .

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

Step 5.2.4.5.11

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

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

Step 5.2.4.5.12

Multiply $6$ by $1$ .

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

Step 5.2.4.5.13

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

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

Step 5.2.4.6

Add $- 12 x^{\frac{2}{3}}$ and $12 x^{\frac{2}{3}}$ .

$f^{- 1} (8 \sqrt[3]{x} - 4) = \frac{8 x + 6 x^{\frac{1}{3}} - 1 + 0 - 6 x^{\frac{1}{3}} + 1}{8}$

Step 5.2.4.7

Add $8 x$ and $0$ .

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

Step 5.2.4.8

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

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

Step 5.2.4.9

Add $8 x$ and $0$ .

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

Step 5.2.4.10

Add $- 1$ and $1$ .

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

Step 5.2.4.11

Add $8 x$ and $0$ .

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

Step 5.2.5

Cancel the common factor of $8$ .

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Step 5.2.5.1

Cancel the common factor.

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

Step 5.2.5.2

Divide $x$ by $1$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.3

Remove parentheses.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

To write $\frac{3 x^{2}}{128}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3}}{512} + \frac{3 x^{2}}{128} \cdot \frac{4}{4} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.2

Write each expression with a common denominator of $512$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.4.2.1

Multiply $\frac{3 x^{2}}{128}$ by $\frac{4}{4}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3}}{512} + \frac{3 x^{2} \cdot 4}{128 \cdot 4} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.2.2

Multiply $128$ by $4$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3}}{512} + \frac{3 x^{2} \cdot 4}{512} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.3

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 3 x^{2} \cdot 4}{512} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.4

Simplify the numerator.

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Step 5.3.4.4.1

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

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Step 5.3.4.4.1.1

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} x + 3 x^{2} \cdot 4}{512} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.4.1.2

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} x + x^{2} (3 \cdot 4)}{512} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.4.1.3

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} (x + 3 \cdot 4)}{512} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.4.2

Multiply $3$ by $4$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} (x + 12)}{512} + \frac{3 x}{32} + \frac{1}{8}} - 4$

Step 5.3.4.5

To write $\frac{3 x}{32}$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} (x + 12)}{512} + \frac{3 x}{32} \cdot \frac{16}{16} + \frac{1}{8}} - 4$

Step 5.3.4.6

Write each expression with a common denominator of $512$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.4.6.1

Multiply $\frac{3 x}{32}$ by $\frac{16}{16}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} (x + 12)}{512} + \frac{3 x \cdot 16}{32 \cdot 16} + \frac{1}{8}} - 4$

Step 5.3.4.6.2

Multiply $32$ by $16$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} (x + 12)}{512} + \frac{3 x \cdot 16}{512} + \frac{1}{8}} - 4$

Step 5.3.4.7

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{2} (x + 12) + 3 x \cdot 16}{512} + \frac{1}{8}} - 4$

Step 5.3.4.8

Simplify the numerator.

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Step 5.3.4.8.1

Factor $x$ out of $x^{2} (x + 12) + 3 x \cdot 16$ .

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Step 5.3.4.8.1.1

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x (x + 12)) + 3 x \cdot 16}{512} + \frac{1}{8}} - 4$

Step 5.3.4.8.1.2

Factor $x$ out of $3 x \cdot 16$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x (x + 12)) + x (3 \cdot 16)}{512} + \frac{1}{8}} - 4$

Step 5.3.4.8.1.3

Factor $x$ out of $x (x (x + 12)) + x (3 \cdot 16)$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x (x + 12) + 3 \cdot 16)}{512} + \frac{1}{8}} - 4$

Step 5.3.4.8.2

Apply the distributive property.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x \cdot x + x \cdot 12 + 3 \cdot 16)}{512} + \frac{1}{8}} - 4$

Step 5.3.4.8.3

Multiply $x$ by $x$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x^{2} + x \cdot 12 + 3 \cdot 16)}{512} + \frac{1}{8}} - 4$

Step 5.3.4.8.4

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x^{2} + 12 \cdot x + 3 \cdot 16)}{512} + \frac{1}{8}} - 4$

Step 5.3.4.8.5

Multiply $3$ by $16$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x^{2} + 12 x + 48)}{512} + \frac{1}{8}} - 4$

Step 5.3.4.9

To write $\frac{1}{8}$ as a fraction with a common denominator, multiply by $\frac{64}{64}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x^{2} + 12 x + 48)}{512} + \frac{1}{8} \cdot \frac{64}{64}} - 4$

Step 5.3.4.10

Write each expression with a common denominator of $512$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.4.10.1

Multiply $\frac{1}{8}$ by $\frac{64}{64}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x^{2} + 12 x + 48)}{512} + \frac{64}{8 \cdot 64}} - 4$

Step 5.3.4.10.2

Multiply $8$ by $64$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x^{2} + 12 x + 48)}{512} + \frac{64}{512}} - 4$

Step 5.3.4.11

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x (x^{2} + 12 x + 48) + 64}{512}} - 4$

Step 5.3.4.12

Simplify the numerator.

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Step 5.3.4.12.1

Apply the distributive property.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x \cdot x^{2} + x (12 x) + x \cdot 48 + 64}{512}} - 4$

Step 5.3.4.12.2

Simplify.

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Step 5.3.4.12.2.1

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

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Step 5.3.4.12.2.1.1

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

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Step 5.3.4.12.2.1.1.1

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x \cdot x^{2} + x (12 x) + x \cdot 48 + 64}{512}} - 4$

Step 5.3.4.12.2.1.1.2

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{1 + 2} + x (12 x) + x \cdot 48 + 64}{512}} - 4$

Step 5.3.4.12.2.1.2

Add $1$ and $2$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + x (12 x) + x \cdot 48 + 64}{512}} - 4$

Step 5.3.4.12.2.2

Rewrite using the commutative property of multiplication.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 12 x \cdot x + x \cdot 48 + 64}{512}} - 4$

Step 5.3.4.12.2.3

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 12 x \cdot x + 48 \cdot x + 64}{512}} - 4$

Step 5.3.4.12.3

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

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Step 5.3.4.12.3.1

Move $x$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 12 (x \cdot x) + 48 \cdot x + 64}{512}} - 4$

Step 5.3.4.12.3.2

Multiply $x$ by $x$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 12 x^{2} + 48 \cdot x + 64}{512}} - 4$

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 12 x^{2} + 48 x + 64}{512}} - 4$

Step 5.3.4.12.4

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

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Step 5.3.4.12.4.1

Regroup terms.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 64 + 12 x^{2} + 48 x}{512}} - 4$

Step 5.3.4.12.4.2

Rewrite $64$ as $4^{3}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{x^{3} + 4^{3} + 12 x^{2} + 48 x}{512}} - 4$

Step 5.3.4.12.4.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 4$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - x \cdot 4 + 4^{2}) + 12 x^{2} + 48 x}{512}} - 4$

Step 5.3.4.12.4.4

Simplify.

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Step 5.3.4.12.4.4.1

Multiply $4$ by $- 1$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - 4 x + 4^{2}) + 12 x^{2} + 48 x}{512}} - 4$

Step 5.3.4.12.4.4.2

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - 4 x + 16) + 12 x^{2} + 48 x}{512}} - 4$

Step 5.3.4.12.4.5

Factor $12 x$ out of $12 x^{2} + 48 x$ .

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Step 5.3.4.12.4.5.1

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - 4 x + 16) + 12 x (x) + 48 x}{512}} - 4$

Step 5.3.4.12.4.5.2

Factor $12 x$ out of $48 x$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - 4 x + 16) + 12 x (x) + 12 x (4)}{512}} - 4$

Step 5.3.4.12.4.5.3

Factor $12 x$ out of $12 x (x) + 12 x (4)$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - 4 x + 16) + 12 x (x + 4)}{512}} - 4$

Step 5.3.4.12.4.6

Factor $x + 4$ out of $(x + 4) (x^{2} - 4 x + 16) + 12 x (x + 4)$ .

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Step 5.3.4.12.4.6.1

Factor $x + 4$ out of $12 x (x + 4)$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - 4 x + 16) + (x + 4) (12 x)}{512}} - 4$

Step 5.3.4.12.4.6.2

Factor $x + 4$ out of $(x + 4) (x^{2} - 4 x + 16) + (x + 4) (12 x)$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} - 4 x + 16 + 12 x)}{512}} - 4$

Step 5.3.4.12.4.7

Add $- 4 x$ and $12 x$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} + 8 x + 16)}{512}} - 4$

Step 5.3.4.12.4.8

Factor using the perfect square rule.

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Step 5.3.4.12.4.8.1

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} + 8 x + 4^{2})}{512}} - 4$

Step 5.3.4.12.4.8.2

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

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

Step 5.3.4.12.4.8.3

Rewrite the polynomial.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) (x^{2} + 2 \cdot x \cdot 4 + 4^{2})}{512}} - 4$

Step 5.3.4.12.4.8.4

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) {(x + 4)}^{2}}{512}} - 4$

Step 5.3.4.13

Multiply $x + 4$ by ${(x + 4)}^{2}$ by adding the exponents.

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Step 5.3.4.13.1

Multiply $x + 4$ by ${(x + 4)}^{2}$ .

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Step 5.3.4.13.1.1

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{(x + 4) {(x + 4)}^{2}}{512}} - 4$

Step 5.3.4.13.1.2

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{{(x + 4)}^{1 + 2}}{512}} - 4$

Step 5.3.4.13.2

Add $1$ and $2$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{{(x + 4)}^{3}}{512}} - 4$

Step 5.3.4.14

Rewrite $512$ as $8^{3}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{\frac{{(x + 4)}^{3}}{8^{3}}} - 4$

Step 5.3.4.15

Rewrite $\frac{{(x + 4)}^{3}}{8^{3}}$ as ${(\frac{x + 4}{8})}^{3}$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 \sqrt[3]{{(\frac{x + 4}{8})}^{3}} - 4$

Step 5.3.4.16

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

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 (\frac{x + 4}{8}) - 4$

Step 5.3.4.17

Cancel the common factor of $8$ .

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Step 5.3.4.17.1

Cancel the common factor.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = 8 (\frac{x + 4}{8}) - 4$

Step 5.3.4.17.2

Rewrite the expression.

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = x + 4 - 4$

Step 5.3.5

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

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Step 5.3.5.1

Subtract $4$ from $4$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$f (\frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}$ is the inverse of $f (x) = 8 \sqrt[3]{x} - 4$ .

$f^{- 1} (x) = \frac{x^{3}}{512} + \frac{3 x^{2}}{128} + \frac{3 x}{32} + \frac{1}{8}$