Find the Inverse cube root of x+5

$\sqrt[3]{x} + 5$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{y} + 5 = x$ .

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

Subtract $5$ from both sides of the equation.

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

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

${\sqrt[3]{y}}^{3} = {(x - 5)}^{3}$

Simplify each side of the equation.

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

${(y^{\frac{1}{3}})}^{3} = {(x - 5)}^{3}$

Simplify the left side.

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

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

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

$y^{\frac{1}{3} \cdot 3} = {(x - 5)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y^{\frac{1}{3} \cdot 3} = {(x - 5)}^{3}$

Rewrite the expression.

$y^{1} = {(x - 5)}^{3}$

Simplify.

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

Simplify the right side.

Simplify ${(x - 5)}^{3}$ .

Use the Binomial Theorem.

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

Simplify each term.

Multiply $- 5$ by $3$ .

$y = x^{3} - 15 x^{2} + 3 x {(- 5)}^{2} + {(- 5)}^{3}$

Raise $- 5$ to the power of $2$ .

$y = x^{3} - 15 x^{2} + 3 x \cdot 25 + {(- 5)}^{3}$

Multiply $25$ by $3$ .

$y = x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}$

Raise $- 5$ to the power of $3$ .

$y = x^{3} - 15 x^{2} + 75 x - 125$

Step 3

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

$f^{- 1} (x) = x^{3} - 15 x^{2} + 75 x - 125$

Step 4

Verify if $f^{- 1} (x) = x^{3} - 15 x^{2} + 75 x - 125$ is the inverse of $f (x) = \sqrt[3]{x} + 5$ .

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

$f^{- 1} (\sqrt[3]{x} + 5) = {(\sqrt[3]{x} + 5)}^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Simplify each term.

Use the Binomial Theorem.

$f^{- 1} (\sqrt[3]{x} + 5) = {\sqrt[3]{x}}^{3} + 3 {\sqrt[3]{x}}^{2} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Simplify each term.

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

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

$f^{- 1} (\sqrt[3]{x} + 5) = {(x^{\frac{1}{3}})}^{3} + 3 {\sqrt[3]{x}}^{2} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x^{\frac{1}{3} \cdot 3} + 3 {\sqrt[3]{x}}^{2} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x^{\frac{3}{3}} + 3 {\sqrt[3]{x}}^{2} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{x} + 5) = x^{\frac{3}{3}} + 3 {\sqrt[3]{x}}^{2} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{x} + 5) = x + 3 {\sqrt[3]{x}}^{2} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Simplify.

$f^{- 1} (\sqrt[3]{x} + 5) = x + 3 {\sqrt[3]{x}}^{2} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 3 \sqrt[3]{x^{2}} \cdot 5 + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Multiply $5$ by $3$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} \cdot 5^{2} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Raise $5$ to the power of $2$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 3 \sqrt[3]{x} \cdot 25 + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Multiply $25$ by $3$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 5^{3} - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

Raise $5$ to the power of $3$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 {(\sqrt[3]{x} + 5)}^{2} + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 ((\sqrt[3]{x} + 5) (\sqrt[3]{x} + 5)) + 75 (\sqrt[3]{x} + 5) - 125$

Expand $(\sqrt[3]{x} + 5) (\sqrt[3]{x} + 5)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x} (\sqrt[3]{x} + 5) + 5 (\sqrt[3]{x} + 5)) + 75 (\sqrt[3]{x} + 5) - 125$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot 5 + 5 (\sqrt[3]{x} + 5)) + 75 (\sqrt[3]{x} + 5) - 125$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot 5 + 5 \sqrt[3]{x} + 5 \cdot 5) + 75 (\sqrt[3]{x} + 5) - 125$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt[3]{x} \sqrt[3]{x}$ .

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot 5 + 5 \sqrt[3]{x} + 5 \cdot 5) + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x} \sqrt[3]{x} + \sqrt[3]{x} \cdot 5 + 5 \sqrt[3]{x} + 5 \cdot 5) + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 ({\sqrt[3]{x}}^{1 + 1} + \sqrt[3]{x} \cdot 5 + 5 \sqrt[3]{x} + 5 \cdot 5) + 75 (\sqrt[3]{x} + 5) - 125$

Add $1$ and $1$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 ({\sqrt[3]{x}}^{2} + \sqrt[3]{x} \cdot 5 + 5 \sqrt[3]{x} + 5 \cdot 5) + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot 5 + 5 \sqrt[3]{x} + 5 \cdot 5) + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x^{2}} + 5 \cdot \sqrt[3]{x} + 5 \sqrt[3]{x} + 5 \cdot 5) + 75 (\sqrt[3]{x} + 5) - 125$

Multiply $5$ by $5$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x^{2}} + 5 \sqrt[3]{x} + 5 \sqrt[3]{x} + 25) + 75 (\sqrt[3]{x} + 5) - 125$

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

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 (\sqrt[3]{x^{2}} + 10 \sqrt[3]{x} + 25) + 75 (\sqrt[3]{x} + 5) - 125$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 \sqrt[3]{x^{2}} - 15 (10 \sqrt[3]{x}) - 15 \cdot 25 + 75 (\sqrt[3]{x} + 5) - 125$

Simplify.

Multiply $10$ by $- 15$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 \sqrt[3]{x^{2}} - 150 \sqrt[3]{x} - 15 \cdot 25 + 75 (\sqrt[3]{x} + 5) - 125$

Multiply $- 15$ by $25$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 \sqrt[3]{x^{2}} - 150 \sqrt[3]{x} - 375 + 75 (\sqrt[3]{x} + 5) - 125$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 \sqrt[3]{x^{2}} - 150 \sqrt[3]{x} - 375 + 75 \sqrt[3]{x} + 75 \cdot 5 - 125$

Multiply $75$ by $5$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 \sqrt[3]{x^{2}} - 150 \sqrt[3]{x} - 375 + 75 \sqrt[3]{x} + 375 - 125$

Simplify by adding terms.

Combine the opposite terms in $x + 15 \sqrt[3]{x^{2}} + 75 \sqrt[3]{x} + 125 - 15 \sqrt[3]{x^{2}} - 150 \sqrt[3]{x} - 375 + 75 \sqrt[3]{x} + 375 - 125$ .

Subtract $15 \sqrt[3]{x^{2}}$ from $15 \sqrt[3]{x^{2}}$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 0 + 75 \sqrt[3]{x} + 125 - 150 \sqrt[3]{x} - 375 + 75 \sqrt[3]{x} + 375 - 125$

Add $x$ and $0$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 75 \sqrt[3]{x} + 125 - 150 \sqrt[3]{x} - 375 + 75 \sqrt[3]{x} + 375 - 125$

Add $- 375$ and $375$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 75 \sqrt[3]{x} + 125 - 150 \sqrt[3]{x} + 0 + 75 \sqrt[3]{x} - 125$

Add $x + 75 \sqrt[3]{x} + 125 - 150 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 75 \sqrt[3]{x} + 125 - 150 \sqrt[3]{x} + 75 \sqrt[3]{x} - 125$

Subtract $125$ from $125$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 75 \sqrt[3]{x} + 0 - 150 \sqrt[3]{x} + 75 \sqrt[3]{x}$

Add $x + 75 \sqrt[3]{x}$ and $0$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x + 75 \sqrt[3]{x} - 150 \sqrt[3]{x} + 75 \sqrt[3]{x}$

Subtract $150 \sqrt[3]{x}$ from $75 \sqrt[3]{x}$ .

$f^{- 1} (\sqrt[3]{x} + 5) = x - 75 \sqrt[3]{x} + 75 \sqrt[3]{x}$

Combine the opposite terms in $x - 75 \sqrt[3]{x} + 75 \sqrt[3]{x}$ .

Add $- 75 \sqrt[3]{x}$ and $75 \sqrt[3]{x}$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (x^{3} - 15 x^{2} + 75 x - 125)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{3} - 15 x^{2} + 75 x - 125) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x - 125} + 5$

Remove parentheses.

$f (x^{3} - 15 x^{2} + 75 x - 125) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x - 125} + 5$

Simplify each term.

Make each term match the terms from the binomial theorem formula.

$f (x^{3} - 15 x^{2} + 75 x - 125) = \sqrt[3]{x^{3} - 3 \cdot (5 x^{2}) + 3 \cdot (5^{2} x) - 1 \cdot 5^{3}} + 5$

Factor $x^{3} - 3 \cdot 5 x^{2} + 3 \cdot 5^{2} x - 1 \cdot 5^{3}$ using the binomial theorem.

$f (x^{3} - 15 x^{2} + 75 x - 125) = \sqrt[3]{{(x - 5)}^{3}} + 5$

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

$f (x^{3} - 15 x^{2} + 75 x - 125) = x - 5 + 5$

Combine the opposite terms in $x - 5 + 5$ .

Add $- 5$ and $5$ .

$f (x^{3} - 15 x^{2} + 75 x - 125) = x + 0$

Add $x$ and $0$ .

$f (x^{3} - 15 x^{2} + 75 x - 125) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{3} - 15 x^{2} + 75 x - 125$ is the inverse of $f (x) = \sqrt[3]{x} + 5$ .

$f^{- 1} (x) = x^{3} - 15 x^{2} + 75 x - 125$

Find the Inverse cube root of x+5

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