Algebra Examples

$f (x) = \frac{{(x + 2)}^{3} - 8}{5}$

Step 1

Write $f (x) = \frac{{(x + 2)}^{3} - 8}{5}$ as an equation.

$y = \frac{{(x + 2)}^{3} - 8}{5}$

Step 2

Interchange the variables.

$x = \frac{{(y + 2)}^{3} - 8}{5}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{{(y + 2)}^{3} - 8}{5} = x$ .

$\frac{{(y + 2)}^{3} - 8}{5} = x$

Step 3.2

Multiply both sides of the equation by $5$ .

$5 \frac{{(y + 2)}^{3} - 8}{5} = 5 x$

Step 3.3

Simplify the left side.

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Step 3.3.1

Cancel the common factor of $5$ .

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Step 3.3.1.1

Cancel the common factor.

$5 \frac{{(y + 2)}^{3} - 8}{5} = 5 x$

Step 3.3.1.2

Rewrite the expression.

${(y + 2)}^{3} - 8 = 5 x$

Step 3.4

Add $8$ to both sides of the equation.

${(y + 2)}^{3} = 5 x + 8$

Step 3.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y + 2 = \sqrt[3]{5 x + 8}$

Step 3.6

Subtract $2$ from both sides of the equation.

$y = \sqrt[3]{5 x + 8} - 2$

Step 4

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

$f^{- 1} (x) = \sqrt[3]{5 x + 8} - 2$

Step 5

Verify if $f^{- 1} (x) = \sqrt[3]{5 x + 8} - 2$ is the inverse of $f (x) = \frac{{(x + 2)}^{3} - 8}{5}$ .

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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 $f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{{(x + 2)}^{3} - 8}{5}) + 8} - 2$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{{(x + 2)}^{3} - 2^{3}}{5}) + 8} - 2$

Step 5.2.3.1.2

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{(x + 2 - 2) ({(x + 2)}^{2} + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3

Simplify.

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Step 5.2.3.1.3.1

Subtract $2$ from $2$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{(x + 0) ({(x + 2)}^{2} + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.2

Add $x$ and $0$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x ({(x + 2)}^{2} + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.3

Rewrite ${(x + 2)}^{2}$ as $(x + 2) (x + 2)$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x ((x + 2) (x + 2) + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.4

Expand $(x + 2) (x + 2)$ using the FOIL Method.

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Step 5.2.3.1.3.4.1

Apply the distributive property.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x (x + 2) + 2 (x + 2) + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.4.2

Apply the distributive property.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x \cdot x + x \cdot 2 + 2 (x + 2) + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.4.3

Apply the distributive property.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2 + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.5

Simplify and combine like terms.

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Step 5.2.3.1.3.5.1

Simplify each term.

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Step 5.2.3.1.3.5.1.1

Multiply $x$ by $x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + x \cdot 2 + 2 x + 2 \cdot 2 + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.5.1.2

Move $2$ to the left of $x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 2 \cdot x + 2 x + 2 \cdot 2 + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.5.1.3

Multiply $2$ by $2$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 2 x + 2 x + 4 + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.5.2

Add $2 x$ and $2 x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 4 x + 4 + (x + 2) \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.6

Apply the distributive property.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 4 x + 4 + x \cdot 2 + 2 \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.7

Move $2$ to the left of $x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 4 x + 4 + 2 \cdot x + 2 \cdot 2 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.8

Multiply $2$ by $2$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 4 x + 4 + 2 \cdot x + 4 + 2^{2})}{5}) + 8} - 2$

Step 5.2.3.1.3.9

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 4 x + 4 + 2 x + 4 + 4)}{5}) + 8} - 2$

Step 5.2.3.1.3.10

Add $4 x$ and $2 x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 6 x + 4 + 4 + 4)}{5}) + 8} - 2$

Step 5.2.3.1.3.11

Add $4$ and $4$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 6 x + 8 + 4)}{5}) + 8} - 2$

Step 5.2.3.1.3.12

Add $8$ and $4$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 6 x + 12)}{5}) + 8} - 2$

Step 5.2.3.2

Cancel the common factor of $5$ .

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Step 5.2.3.2.1

Cancel the common factor.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{5 (\frac{x (x^{2} + 6 x + 12)}{5}) + 8} - 2$

Step 5.2.3.2.2

Rewrite the expression.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x (x^{2} + 6 x + 12) + 8} - 2$

Step 5.2.3.3

Apply the distributive property.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x \cdot x^{2} + x (6 x) + x \cdot 12 + 8} - 2$

Step 5.2.3.4

Simplify.

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Step 5.2.3.4.1

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

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Step 5.2.3.4.1.1

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

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Step 5.2.3.4.1.1.1

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x \cdot x^{2} + x (6 x) + x \cdot 12 + 8} - 2$

Step 5.2.3.4.1.1.2

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{1 + 2} + x (6 x) + x \cdot 12 + 8} - 2$

Step 5.2.3.4.1.2

Add $1$ and $2$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + x (6 x) + x \cdot 12 + 8} - 2$

Step 5.2.3.4.2

Rewrite using the commutative property of multiplication.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + 6 x \cdot x + x \cdot 12 + 8} - 2$

Step 5.2.3.4.3

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + 6 x \cdot x + 12 \cdot x + 8} - 2$

Step 5.2.3.5

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

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Step 5.2.3.5.1

Move $x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + 6 (x \cdot x) + 12 \cdot x + 8} - 2$

Step 5.2.3.5.2

Multiply $x$ by $x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + 6 x^{2} + 12 \cdot x + 8} - 2$

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + 6 x^{2} + 12 x + 8} - 2$

Step 5.2.3.6

Rewrite $x^{3} + 6 x^{2} + 12 x + 8$ in a factored form.

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Step 5.2.3.6.1

Regroup terms.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + 8 + 6 x^{2} + 12 x} - 2$

Step 5.2.3.6.2

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{x^{3} + 2^{3} + 6 x^{2} + 12 x} - 2$

Step 5.2.3.6.3

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 $b = 2$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - x \cdot 2 + 2^{2}) + 6 x^{2} + 12 x} - 2$

Step 5.2.3.6.4

Simplify.

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Step 5.2.3.6.4.1

Multiply $2$ by $- 1$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - 2 x + 2^{2}) + 6 x^{2} + 12 x} - 2$

Step 5.2.3.6.4.2

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - 2 x + 4) + 6 x^{2} + 12 x} - 2$

Step 5.2.3.6.5

Factor $6 x$ out of $6 x^{2} + 12 x$ .

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Step 5.2.3.6.5.1

Factor $6 x$ out of $6 x^{2}$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - 2 x + 4) + 6 x (x) + 12 x} - 2$

Step 5.2.3.6.5.2

Factor $6 x$ out of $12 x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - 2 x + 4) + 6 x (x) + 6 x (2)} - 2$

Step 5.2.3.6.5.3

Factor $6 x$ out of $6 x (x) + 6 x (2)$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - 2 x + 4) + 6 x (x + 2)} - 2$

Step 5.2.3.6.6

Factor $x + 2$ out of $(x + 2) (x^{2} - 2 x + 4) + 6 x (x + 2)$ .

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Step 5.2.3.6.6.1

Factor $x + 2$ out of $6 x (x + 2)$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - 2 x + 4) + (x + 2) (6 x)} - 2$

Step 5.2.3.6.6.2

Factor $x + 2$ out of $(x + 2) (x^{2} - 2 x + 4) + (x + 2) (6 x)$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} - 2 x + 4 + 6 x)} - 2$

Step 5.2.3.6.7

Add $- 2 x$ and $6 x$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} + 4 x + 4)} - 2$

Step 5.2.3.6.8

Factor using the perfect square rule.

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Step 5.2.3.6.8.1

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} + 4 x + 2^{2})} - 2$

Step 5.2.3.6.8.2

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

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

Step 5.2.3.6.8.3

Rewrite the polynomial.

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) (x^{2} + 2 \cdot x \cdot 2 + 2^{2})} - 2$

Step 5.2.3.6.8.4

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) {(x + 2)}^{2}} - 2$

Step 5.2.3.7

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

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Step 5.2.3.7.1

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

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Step 5.2.3.7.1.1

Raise $x + 2$ to the power of $1$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{(x + 2) {(x + 2)}^{2}} - 2$

Step 5.2.3.7.1.2

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{{(x + 2)}^{1 + 2}} - 2$

Step 5.2.3.7.2

Add $1$ and $2$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = \sqrt[3]{{(x + 2)}^{3}} - 2$

Step 5.2.3.8

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

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = x + 2 - 2$

Step 5.2.4

Combine the opposite terms in $x + 2 - 2$ .

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Step 5.2.4.1

Subtract $2$ from $2$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\frac{{(x + 2)}^{3} - 8}{5}) = x$

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 $f (\sqrt[3]{5 x + 8} - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{{((\sqrt[3]{5 x + 8} - 2) + 2)}^{3} - 8}{5}$

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

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

$f (\sqrt[3]{5 x + 8} - 2) = \frac{{(\sqrt[3]{5 x + 8} - 2 + 2)}^{3} - 2^{3}}{5}$

Step 5.3.3.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = \sqrt[3]{5 x + 8} - 2 + 2$ and $b = 2$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2 + 2 - 2) ({(\sqrt[3]{5 x + 8} - 2 + 2)}^{2} + (\sqrt[3]{5 x + 8} - 2 + 2) \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3

Simplify.

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Step 5.3.3.3.1

Add $- 2$ and $2$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} + 0 - 2) ({(\sqrt[3]{5 x + 8} - 2 + 2)}^{2} + (\sqrt[3]{5 x + 8} - 2 + 2) \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3.2

Add $\sqrt[3]{5 x + 8}$ and $0$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) ({(\sqrt[3]{5 x + 8} - 2 + 2)}^{2} + (\sqrt[3]{5 x + 8} - 2 + 2) \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3.3

Combine the opposite terms in $\sqrt[3]{5 x + 8} - 2 + 2$ .

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Step 5.3.3.3.3.1

Add $- 2$ and $2$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) ({(\sqrt[3]{5 x + 8} + 0)}^{2} + (\sqrt[3]{5 x + 8} - 2 + 2) \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3.3.2

Add $\sqrt[3]{5 x + 8}$ and $0$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) ({\sqrt[3]{5 x + 8}}^{2} + (\sqrt[3]{5 x + 8} - 2 + 2) \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3.4

Rewrite ${\sqrt[3]{5 x + 8}}^{2}$ as $\sqrt[3]{{(5 x + 8)}^{2}}$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) (\sqrt[3]{{(5 x + 8)}^{2}} + (\sqrt[3]{5 x + 8} - 2 + 2) \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3.5

Combine the opposite terms in $\sqrt[3]{5 x + 8} - 2 + 2$ .

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Step 5.3.3.3.5.1

Add $- 2$ and $2$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) (\sqrt[3]{{(5 x + 8)}^{2}} + (\sqrt[3]{5 x + 8} + 0) \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3.5.2

Add $\sqrt[3]{5 x + 8}$ and $0$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) (\sqrt[3]{{(5 x + 8)}^{2}} + \sqrt[3]{5 x + 8} \cdot 2 + 2^{2})}{5}$

Step 5.3.3.3.6

Move $2$ to the left of $\sqrt[3]{5 x + 8}$ .

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) (\sqrt[3]{{(5 x + 8)}^{2}} + 2 \cdot \sqrt[3]{5 x + 8} + 2^{2})}{5}$

Step 5.3.3.3.7

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

$f (\sqrt[3]{5 x + 8} - 2) = \frac{(\sqrt[3]{5 x + 8} - 2) (\sqrt[3]{{(5 x + 8)}^{2}} + 2 \sqrt[3]{5 x + 8} + 4)}{5}$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{5 x + 8} - 2$ is the inverse of $f (x) = \frac{{(x + 2)}^{3} - 8}{5}$ .

$f^{- 1} (x) = \sqrt[3]{5 x + 8} - 2$