Find the Inverse y=4+ cube root of x-4

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Subtract $4$ from both sides of the equation.

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

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

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{y - 4}$ as ${(y - 4)}^{\frac{1}{3}}$ .

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

Simplify the left side.

Simplify ${({(y - 4)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(y - 4)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y - 4 = {(x - 4)}^{3}$

Simplify the right side.

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

Use the Binomial Theorem.

$y - 4 = x^{3} + 3 x^{2} \cdot - 4 + 3 x {(- 4)}^{2} + {(- 4)}^{3}$

Simplify each term.

Multiply $- 4$ by $3$ .

$y - 4 = x^{3} - 12 x^{2} + 3 x {(- 4)}^{2} + {(- 4)}^{3}$

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

$y - 4 = x^{3} - 12 x^{2} + 3 x \cdot 16 + {(- 4)}^{3}$

Multiply $16$ by $3$ .

$y - 4 = x^{3} - 12 x^{2} + 48 x + {(- 4)}^{3}$

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

$y - 4 = x^{3} - 12 x^{2} + 48 x - 64$

Move all terms not containing $y$ to the right side of the equation.

Add $4$ to both sides of the equation.

$y = x^{3} - 12 x^{2} + 48 x - 64 + 4$

Add $- 64$ and $4$ .

$y = x^{3} - 12 x^{2} + 48 x - 60$

Step 3

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

$f^{- 1} (x) = x^{3} - 12 x^{2} + 48 x - 60$

Step 4

Verify if $f^{- 1} (x) = x^{3} - 12 x^{2} + 48 x - 60$ is the inverse of $f (x) = 4 + \sqrt[3]{x - 4}$ .

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

$f^{- 1} (4 + \sqrt[3]{x - 4}) = {(4 + \sqrt[3]{x - 4})}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Simplify each term.

Use the Binomial Theorem.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 4^{3} + 3 \cdot (4^{2} \sqrt[3]{x - 4}) + 3 \cdot (4 {\sqrt[3]{x - 4}}^{2}) + {\sqrt[3]{x - 4}}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Simplify each term.

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

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 3 \cdot (4^{2} \sqrt[3]{x - 4}) + 3 \cdot (4 {\sqrt[3]{x - 4}}^{2}) + {\sqrt[3]{x - 4}}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

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

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 3 \cdot (16 \sqrt[3]{x - 4}) + 3 \cdot (4 {\sqrt[3]{x - 4}}^{2}) + {\sqrt[3]{x - 4}}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Multiply $3$ by $16$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 3 \cdot (4 {\sqrt[3]{x - 4}}^{2}) + {\sqrt[3]{x - 4}}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Multiply $3$ by $4$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 {\sqrt[3]{x - 4}}^{2} + {\sqrt[3]{x - 4}}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Rewrite ${\sqrt[3]{x - 4}}^{2}$ as ${({(x - 4)}^{2})}^{\frac{1}{3}}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + {\sqrt[3]{x - 4}}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Rewrite ${\sqrt[3]{x - 4}}^{3}$ as $x - 4$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 4}$ as ${(x - 4)}^{\frac{1}{3}}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + {({(x - 4)}^{\frac{1}{3}})}^{3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

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

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + {(x - 4)}^{\frac{1}{3} \cdot 3} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

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

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + {(x - 4)}^{\frac{3}{3}} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + {(x - 4)}^{\frac{3}{3}} - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Rewrite the expression.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 4 - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Simplify.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 64 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 4 - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Subtract $4$ from $64$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 {(4 + \sqrt[3]{x - 4})}^{2} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Rewrite ${(4 + \sqrt[3]{x - 4})}^{2}$ as $(4 + \sqrt[3]{x - 4}) (4 + \sqrt[3]{x - 4})$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 ((4 + \sqrt[3]{x - 4}) (4 + \sqrt[3]{x - 4})) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Expand $(4 + \sqrt[3]{x - 4}) (4 + \sqrt[3]{x - 4})$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (4 (4 + \sqrt[3]{x - 4}) + \sqrt[3]{x - 4} (4 + \sqrt[3]{x - 4})) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Apply the distributive property.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (4 \cdot 4 + 4 \sqrt[3]{x - 4} + \sqrt[3]{x - 4} (4 + \sqrt[3]{x - 4})) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Apply the distributive property.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (4 \cdot 4 + 4 \sqrt[3]{x - 4} + \sqrt[3]{x - 4} \cdot 4 + \sqrt[3]{x - 4} \sqrt[3]{x - 4}) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Simplify and combine like terms.

Simplify each term.

Multiply $4$ by $4$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (16 + 4 \sqrt[3]{x - 4} + \sqrt[3]{x - 4} \cdot 4 + \sqrt[3]{x - 4} \sqrt[3]{x - 4}) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Move $4$ to the left of $\sqrt[3]{x - 4}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (16 + 4 \sqrt[3]{x - 4} + 4 \sqrt[3]{x - 4} + \sqrt[3]{x - 4} \sqrt[3]{x - 4}) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Multiply $\sqrt[3]{x - 4} \sqrt[3]{x - 4}$ .

Raise $\sqrt[3]{x - 4}$ to the power of $1$ .

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

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (16 + 4 \sqrt[3]{x - 4} + 4 \sqrt[3]{x - 4} + {\sqrt[3]{x - 4}}^{1 + 1}) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Add $1$ and $1$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (16 + 4 \sqrt[3]{x - 4} + 4 \sqrt[3]{x - 4} + {\sqrt[3]{x - 4}}^{2}) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Rewrite ${\sqrt[3]{x - 4}}^{2}$ as ${({(x - 4)}^{2})}^{\frac{1}{3}}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (16 + 4 \sqrt[3]{x - 4} + 4 \sqrt[3]{x - 4} + \sqrt[3]{{(x - 4)}^{2}}) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Add $4 \sqrt[3]{x - 4}$ and $4 \sqrt[3]{x - 4}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 (16 + 8 \sqrt[3]{x - 4} + \sqrt[3]{{(x - 4)}^{2}}) + 48 (4 + \sqrt[3]{x - 4}) - 60$

Apply the distributive property.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 12 \cdot 16 - 12 (8 \sqrt[3]{x - 4}) - 12 \sqrt[3]{{(x - 4)}^{2}} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Simplify.

Multiply $- 12$ by $16$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 192 - 12 (8 \sqrt[3]{x - 4}) - 12 \sqrt[3]{{(x - 4)}^{2}} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Multiply $8$ by $- 12$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 192 - 96 \sqrt[3]{x - 4} - 12 \sqrt[3]{{(x - 4)}^{2}} + 48 (4 + \sqrt[3]{x - 4}) - 60$

Apply the distributive property.

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 192 - 96 \sqrt[3]{x - 4} - 12 \sqrt[3]{{(x - 4)}^{2}} + 48 \cdot 4 + 48 \sqrt[3]{x - 4} - 60$

Multiply $48$ by $4$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 192 - 96 \sqrt[3]{x - 4} - 12 \sqrt[3]{{(x - 4)}^{2}} + 192 + 48 \sqrt[3]{x - 4} - 60$

Simplify by adding terms.

Combine the opposite terms in $60 + 48 \sqrt[3]{x - 4} + 12 \sqrt[3]{{(x - 4)}^{2}} + x - 192 - 96 \sqrt[3]{x - 4} - 12 \sqrt[3]{{(x - 4)}^{2}} + 192 + 48 \sqrt[3]{x - 4} - 60$ .

Subtract $12 \sqrt[3]{{(x - 4)}^{2}}$ from $12 \sqrt[3]{{(x - 4)}^{2}}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + 0 + x - 192 - 96 \sqrt[3]{x - 4} + 192 + 48 \sqrt[3]{x - 4} - 60$

Add $60 + 48 \sqrt[3]{x - 4}$ and $0$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + x - 192 - 96 \sqrt[3]{x - 4} + 192 + 48 \sqrt[3]{x - 4} - 60$

Add $- 192$ and $192$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + x + 0 - 96 \sqrt[3]{x - 4} + 48 \sqrt[3]{x - 4} - 60$

Add $60 + 48 \sqrt[3]{x - 4} + x$ and $0$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 60 + 48 \sqrt[3]{x - 4} + x - 96 \sqrt[3]{x - 4} + 48 \sqrt[3]{x - 4} - 60$

Subtract $60$ from $60$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 0 + 48 \sqrt[3]{x - 4} + x - 96 \sqrt[3]{x - 4} + 48 \sqrt[3]{x - 4}$

Add $0$ and $48 \sqrt[3]{x - 4}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = 48 \sqrt[3]{x - 4} + x - 96 \sqrt[3]{x - 4} + 48 \sqrt[3]{x - 4}$

Subtract $96 \sqrt[3]{x - 4}$ from $48 \sqrt[3]{x - 4}$ .

$f^{- 1} (4 + \sqrt[3]{x - 4}) = x - 48 \sqrt[3]{x - 4} + 48 \sqrt[3]{x - 4}$

Combine the opposite terms in $x - 48 \sqrt[3]{x - 4} + 48 \sqrt[3]{x - 4}$ .

Add $- 48 \sqrt[3]{x - 4}$ and $48 \sqrt[3]{x - 4}$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (x^{3} - 12 x^{2} + 48 x - 60)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{(x^{3} - 12 x^{2} + 48 x - 60) - 4}$

Simplify each term.

Subtract $4$ from $- 60$ .

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{x^{3} - 12 x^{2} + 48 x - 64}$

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

Factor $x^{3} - 12 x^{2} + 48 x - 64$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 64, \pm 2, \pm 32, \pm 4, \pm 16, \pm 8 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 64, \pm 2, \pm 32, \pm 4, \pm 16, \pm 8$

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

Substitute $4$ into the polynomial.

$4^{3} - 12 \cdot 4^{2} + 48 \cdot 4 - 64$

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

$64 - 12 \cdot 4^{2} + 48 \cdot 4 - 64$

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

$64 - 12 \cdot 16 + 48 \cdot 4 - 64$

Multiply $- 12$ by $16$ .

$64 - 192 + 48 \cdot 4 - 64$

Subtract $192$ from $64$ .

$- 128 + 48 \cdot 4 - 64$

Multiply $48$ by $4$ .

$- 128 + 192 - 64$

Add $- 128$ and $192$ .

$64 - 64$

Subtract $64$ from $64$ .

$0$

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 12 x^{2} + 48 x - 64}{x - 4}$

Divide $x^{3} - 12 x^{2} + 48 x - 64$ by $x - 4$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$4$

$x^{3}$

$12 x^{2}$

$48 x$

$64$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			+	$x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 4 x^{2}$

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$

Divide the highest order term in the dividend $- 8 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					-	$8 x^{2}$	+	$32 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 x^{2} + 32 x$

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$8 x$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$

Divide the highest order term in the dividend $16 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$
							+	$16 x$	-	$64$

The expression needs to be subtracted from the dividend, so change all the signs in $16 x - 64$

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$
							-	$16 x$	+	$64$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$4$		$x^{3}$	-	$12 x^{2}$	+	$48 x$	-	$64$
			-	$x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$48 x$
					+	$8 x^{2}$	-	$32 x$
							+	$16 x$	-	$64$
							-	$16 x$	+	$64$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - 8 x + 16$

Write $x^{3} - 12 x^{2} + 48 x - 64$ as a set of factors.

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{(x - 4) (x^{2} - 8 x + 16)}$

Factor using the perfect square rule.

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

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{(x - 4) (x^{2} - 8 x + 4^{2})}$

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

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

Rewrite the polynomial.

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{(x - 4) (x^{2} - 2 \cdot x \cdot 4 + 4^{2})}$

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

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{(x - 4) {(x - 4)}^{2}}$

Combine like factors.

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

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{(x - 4) {(x - 4)}^{2}}$

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

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{{(x - 4)}^{1 + 2}}$

Add $1$ and $2$ .

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + \sqrt[3]{{(x - 4)}^{3}}$

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

$f (x^{3} - 12 x^{2} + 48 x - 60) = 4 + x - 4$

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

Subtract $4$ from $4$ .

$f (x^{3} - 12 x^{2} + 48 x - 60) = x + 0$

Add $x$ and $0$ .

$f (x^{3} - 12 x^{2} + 48 x - 60) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{3} - 12 x^{2} + 48 x - 60$ is the inverse of $f (x) = 4 + \sqrt[3]{x - 4}$ .

$f^{- 1} (x) = x^{3} - 12 x^{2} + 48 x - 60$

Find the Inverse y=4+ cube root of x-4

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