Algebra Examples

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

Step 1

Write $f (x) = \sqrt[3]{1 - x^{3}}$ as an equation.

$y = \sqrt[3]{1 - x^{3}}$

Step 2

Interchange the variables.

$x = \sqrt[3]{1 - y^{3}}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\sqrt[3]{1 - y^{3}} = x$ .

$\sqrt[3]{1 - y^{3}} = x$

Step 3.2

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

${\sqrt[3]{1 - y^{3}}}^{3} = x^{3}$

Step 3.3

Simplify each side of the equation.

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

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

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Step 3.3.2.1.1.1

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

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

Step 3.3.2.1.1.2

Cancel the common factor of $3$ .

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Step 3.3.2.1.1.2.1

Cancel the common factor.

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

Step 3.3.2.1.1.2.2

Rewrite the expression.

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

Step 3.3.2.1.2

Simplify.

$1 - y^{3} = x^{3}$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Subtract $1$ from both sides of the equation.

$- y^{3} = x^{3} - 1$

Step 3.4.2

Divide each term in $- y^{3} = x^{3} - 1$ by $- 1$ and simplify.

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Step 3.4.2.1

Divide each term in $- y^{3} = x^{3} - 1$ by $- 1$ .

$\frac{- y^{3}}{- 1} = \frac{x^{3}}{- 1} + \frac{- 1}{- 1}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{y^{3}}{1} = \frac{x^{3}}{- 1} + \frac{- 1}{- 1}$

Step 3.4.2.2.2

Divide $y^{3}$ by $1$ .

$y^{3} = \frac{x^{3}}{- 1} + \frac{- 1}{- 1}$

Step 3.4.2.3

Simplify the right side.

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Step 3.4.2.3.1

Simplify each term.

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Step 3.4.2.3.1.1

Move the negative one from the denominator of $\frac{x^{3}}{- 1}$ .

$y^{3} = - 1 \cdot x^{3} + \frac{- 1}{- 1}$

Step 3.4.2.3.1.2

Rewrite $- 1 \cdot x^{3}$ as $- x^{3}$ .

$y^{3} = - x^{3} + \frac{- 1}{- 1}$

Step 3.4.2.3.1.3

Divide $- 1$ by $- 1$ .

$y^{3} = - x^{3} + 1$

Step 3.4.3

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

$y = \sqrt[3]{- x^{3} + 1}$

Step 3.4.4

Simplify $\sqrt[3]{- x^{3} + 1}$ .

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Step 3.4.4.1

Simplify with factoring out.

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Step 3.4.4.1.1

Rewrite $- x^{3}$ as ${(- x)}^{3}$ .

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

Step 3.4.4.1.2

Rewrite $1$ as $1^{3}$ .

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

Step 3.4.4.2

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 $b = 1$ .

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

Step 3.4.4.3

Simplify.

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Step 3.4.4.3.1

Apply the product rule to $- x$ .

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

Step 3.4.4.3.2

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

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

Step 3.4.4.3.3

Multiply $x^{2}$ by $1$ .

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

Step 3.4.4.3.4

Multiply $- - x$ .

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Step 3.4.4.3.4.1

Multiply $- 1$ by $- 1$ .

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

Step 3.4.4.3.4.2

Multiply $x$ by $1$ .

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

Step 3.4.4.3.5

Multiply $x$ by $1$ .

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

Step 3.4.4.3.6

One to any power is one.

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify the expression.

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Step 5.2.3.1

Remove parentheses.

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

Step 5.2.3.2

Rewrite $1$ as $1^{3}$ .

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

Step 5.2.4

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 $a = 1$ and $b = x$ .

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

Step 5.2.5

Simplify.

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Step 5.2.5.1

One to any power is one.

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

Step 5.2.5.2

Multiply $x$ by $1$ .

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

Step 5.2.6

Rewrite $1$ as $1^{3}$ .

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

Step 5.2.7

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 $a = 1$ and $b = x$ .

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

Step 5.2.8

Simplify.

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Step 5.2.8.1

One to any power is one.

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

Step 5.2.8.2

Multiply $x$ by $1$ .

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

Step 5.2.9

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

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

Step 5.2.10

Simplify the expression.

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Step 5.2.10.1

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

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

Step 5.2.10.2

Rewrite $1$ as $1^{3}$ .

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

Step 5.2.11

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 $a = 1$ and $b = x$ .

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

Step 5.2.12

Simplify.

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Step 5.2.12.1

One to any power is one.

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

Step 5.2.12.2

Multiply $x$ by $1$ .

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

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

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

Step 5.3.3

Rewrite $1$ as $1^{3}$ .

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

Step 5.3.4

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 $a = 1$ and $b = \sqrt[3]{(- x + 1) (x^{2} + x + 1)}$ .

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

Step 5.3.5

Simplify.

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Step 5.3.5.1

One to any power is one.

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

Step 5.3.5.2

Multiply $\sqrt[3]{(- x + 1) (x^{2} + x + 1)}$ by $1$ .

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

Step 5.3.5.3

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

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

Step 5.3.5.4

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

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

Step 5.4

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

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