Find the Inverse cube root of x-8

$\sqrt[3]{x - 8}$

Interchange the variables.

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

Solve for $y$ .

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

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

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

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

Simplify each side of the equation.

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

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

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

Cancel the common factor of $3$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y - 8 = x^{3}$

Add $8$ to both sides of the equation.

$y = x^{3} + 8$

Solve for $y$ and replace with $f^{- 1} (x)$ .

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

$f^{- 1} (x) = x^{3} + 8$

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

$g (\sqrt[3]{x - 8}) = {(\sqrt[3]{x - 8})}^{3} + 8$

Rewrite ${\sqrt[3]{x - 8}}^{3}$ as $x - 8$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

$g (\sqrt[3]{x - 8}) = (x - 8) + 8$

Simplify.

$g (\sqrt[3]{x - 8}) = x - 8 + 8$

Combine the opposite terms in $x - 8 + 8$ .

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Add $- 8$ and $8$ .

$g (\sqrt[3]{x - 8}) = x + 0$

Add $x$ and $0$ .

$g (\sqrt[3]{x - 8}) = x$

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

$f^{- 1} (x) = x^{3} + 8$

Find the Inverse cube root of x-8

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