Find the Inverse f(x) = cube root of 9+x+3

$f (x) = \sqrt[3]{9 + x} + 3$

Step 1

Write $f (x) = \sqrt[3]{9 + x} + 3$ as an equation.

$y = \sqrt[3]{9 + x} + 3$

Step 2

Interchange the variables.

$x = \sqrt[3]{9 + y} + 3$

Step 3

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{9 + y} + 3 = x$ .

$\sqrt[3]{9 + y} + 3 = x$

Subtract $3$ from both sides of the equation.

$\sqrt[3]{9 + y} = x - 3$

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

${\sqrt[3]{9 + y}}^{3} = {(x - 3)}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{9 + y}$ as ${(9 + y)}^{\frac{1}{3}}$ .

${({(9 + y)}^{\frac{1}{3}})}^{3} = {(x - 3)}^{3}$

Simplify the left side.

Simplify ${({(9 + y)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(9 + y)}^{\frac{1}{3}})}^{3}$ .

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

${(9 + y)}^{\frac{1}{3} \cdot 3} = {(x - 3)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(9 + y)}^{\frac{1}{3} \cdot 3} = {(x - 3)}^{3}$

Rewrite the expression.

${(9 + y)}^{1} = {(x - 3)}^{3}$

Simplify.

$9 + y = {(x - 3)}^{3}$

Simplify the right side.

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

Use the Binomial Theorem.

$9 + y = x^{3} + 3 x^{2} \cdot - 3 + 3 x {(- 3)}^{2} + {(- 3)}^{3}$

Simplify each term.

Multiply $- 3$ by $3$ .

$9 + y = x^{3} - 9 x^{2} + 3 x {(- 3)}^{2} + {(- 3)}^{3}$

Raise $- 3$ to the power of $2$ .

$9 + y = x^{3} - 9 x^{2} + 3 x \cdot 9 + {(- 3)}^{3}$

Multiply $9$ by $3$ .

$9 + y = x^{3} - 9 x^{2} + 27 x + {(- 3)}^{3}$

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

$9 + y = x^{3} - 9 x^{2} + 27 x - 27$

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

Subtract $9$ from both sides of the equation.

$y = x^{3} - 9 x^{2} + 27 x - 27 - 9$

Subtract $9$ from $- 27$ .

$y = x^{3} - 9 x^{2} + 27 x - 36$

Step 4

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

$f^{- 1} (x) = x^{3} - 9 x^{2} + 27 x - 36$

Step 5

Verify if $f^{- 1} (x) = x^{3} - 9 x^{2} + 27 x - 36$ is the inverse of $f (x) = \sqrt[3]{9 + x} + 3$ .

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 $f^{- 1} (\sqrt[3]{9 + x} + 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = {(\sqrt[3]{9 + x} + 3)}^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Simplify each term.

Use the Binomial Theorem.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = {\sqrt[3]{9 + x}}^{3} + 3 {\sqrt[3]{9 + x}}^{2} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Simplify each term.

Rewrite ${\sqrt[3]{9 + x}}^{3}$ as $9 + x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{9 + x}$ as ${(9 + x)}^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = {({(9 + x)}^{\frac{1}{3}})}^{3} + 3 {\sqrt[3]{9 + x}}^{2} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

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

$f^{- 1} (\sqrt[3]{9 + x} + 3) = {(9 + x)}^{\frac{1}{3} \cdot 3} + 3 {\sqrt[3]{9 + x}}^{2} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

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

$f^{- 1} (\sqrt[3]{9 + x} + 3) = {(9 + x)}^{\frac{3}{3}} + 3 {\sqrt[3]{9 + x}}^{2} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = {(9 + x)}^{\frac{3}{3}} + 3 {\sqrt[3]{9 + x}}^{2} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = (9 + x) + 3 {\sqrt[3]{9 + x}}^{2} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Simplify.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 3 {\sqrt[3]{9 + x}}^{2} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Rewrite ${\sqrt[3]{9 + x}}^{2}$ as ${({(9 + x)}^{2})}^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 3 \sqrt[3]{{(9 + x)}^{2}} \cdot 3 + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Multiply $3$ by $3$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 3 \sqrt[3]{9 + x} \cdot 3^{2} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Multiply $3$ by $3^{2}$ by adding the exponents.

Move $3^{2}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 3^{2} \cdot (3 \sqrt[3]{9 + x}) + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Multiply $3^{2}$ by $3$ .

Raise $3$ to the power of $1$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 3^{2} \cdot (3 \sqrt[3]{9 + x}) + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

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

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 3^{2 + 1} \sqrt[3]{9 + x} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Add $2$ and $1$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 3^{3} \sqrt[3]{9 + x} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

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

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} + 3^{3} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

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

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 9 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} + 27 - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Add $9$ and $27$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 {(\sqrt[3]{9 + x} + 3)}^{2} + 27 (\sqrt[3]{9 + x} + 3) - 36$

Rewrite ${(\sqrt[3]{9 + x} + 3)}^{2}$ as $(\sqrt[3]{9 + x} + 3) (\sqrt[3]{9 + x} + 3)$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 ((\sqrt[3]{9 + x} + 3) (\sqrt[3]{9 + x} + 3)) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Expand $(\sqrt[3]{9 + x} + 3) (\sqrt[3]{9 + x} + 3)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 (\sqrt[3]{9 + x} (\sqrt[3]{9 + x} + 3) + 3 (\sqrt[3]{9 + x} + 3)) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 (\sqrt[3]{9 + x} \sqrt[3]{9 + x} + \sqrt[3]{9 + x} \cdot 3 + 3 (\sqrt[3]{9 + x} + 3)) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 (\sqrt[3]{9 + x} \sqrt[3]{9 + x} + \sqrt[3]{9 + x} \cdot 3 + 3 \sqrt[3]{9 + x} + 3 \cdot 3) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt[3]{9 + x} \sqrt[3]{9 + x}$ .

Raise $\sqrt[3]{9 + x}$ to the power of $1$ .

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

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 ({\sqrt[3]{9 + x}}^{1 + 1} + \sqrt[3]{9 + x} \cdot 3 + 3 \sqrt[3]{9 + x} + 3 \cdot 3) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Add $1$ and $1$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 ({\sqrt[3]{9 + x}}^{2} + \sqrt[3]{9 + x} \cdot 3 + 3 \sqrt[3]{9 + x} + 3 \cdot 3) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Rewrite ${\sqrt[3]{9 + x}}^{2}$ as ${({(9 + x)}^{2})}^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 (\sqrt[3]{{(9 + x)}^{2}} + \sqrt[3]{9 + x} \cdot 3 + 3 \sqrt[3]{9 + x} + 3 \cdot 3) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Move $3$ to the left of $\sqrt[3]{9 + x}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 (\sqrt[3]{{(9 + x)}^{2}} + 3 \cdot \sqrt[3]{9 + x} + 3 \sqrt[3]{9 + x} + 3 \cdot 3) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Multiply $3$ by $3$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 (\sqrt[3]{{(9 + x)}^{2}} + 3 \sqrt[3]{9 + x} + 3 \sqrt[3]{9 + x} + 9) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Add $3 \sqrt[3]{9 + x}$ and $3 \sqrt[3]{9 + x}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 (\sqrt[3]{{(9 + x)}^{2}} + 6 \sqrt[3]{9 + x} + 9) + 27 (\sqrt[3]{9 + x} + 3) - 36$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 \sqrt[3]{{(9 + x)}^{2}} - 9 (6 \sqrt[3]{9 + x}) - 9 \cdot 9 + 27 (\sqrt[3]{9 + x} + 3) - 36$

Simplify.

Multiply $6$ by $- 9$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 \sqrt[3]{{(9 + x)}^{2}} - 54 \sqrt[3]{9 + x} - 9 \cdot 9 + 27 (\sqrt[3]{9 + x} + 3) - 36$

Multiply $- 9$ by $9$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 \sqrt[3]{{(9 + x)}^{2}} - 54 \sqrt[3]{9 + x} - 81 + 27 (\sqrt[3]{9 + x} + 3) - 36$

Apply the distributive property.

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 \sqrt[3]{{(9 + x)}^{2}} - 54 \sqrt[3]{9 + x} - 81 + 27 \sqrt[3]{9 + x} + 27 \cdot 3 - 36$

Multiply $27$ by $3$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 \sqrt[3]{{(9 + x)}^{2}} - 54 \sqrt[3]{9 + x} - 81 + 27 \sqrt[3]{9 + x} + 81 - 36$

Simplify by adding terms.

Combine the opposite terms in $36 + x + 9 \sqrt[3]{{(9 + x)}^{2}} + 27 \sqrt[3]{9 + x} - 9 \sqrt[3]{{(9 + x)}^{2}} - 54 \sqrt[3]{9 + x} - 81 + 27 \sqrt[3]{9 + x} + 81 - 36$ .

Subtract $9 \sqrt[3]{{(9 + x)}^{2}}$ from $9 \sqrt[3]{{(9 + x)}^{2}}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 0 + 27 \sqrt[3]{9 + x} - 54 \sqrt[3]{9 + x} - 81 + 27 \sqrt[3]{9 + x} + 81 - 36$

Add $36 + x$ and $0$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 27 \sqrt[3]{9 + x} - 54 \sqrt[3]{9 + x} - 81 + 27 \sqrt[3]{9 + x} + 81 - 36$

Add $- 81$ and $81$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 27 \sqrt[3]{9 + x} - 54 \sqrt[3]{9 + x} + 0 + 27 \sqrt[3]{9 + x} - 36$

Add $36 + x + 27 \sqrt[3]{9 + x} - 54 \sqrt[3]{9 + x}$ and $0$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 36 + x + 27 \sqrt[3]{9 + x} - 54 \sqrt[3]{9 + x} + 27 \sqrt[3]{9 + x} - 36$

Subtract $36$ from $36$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = 0 + x + 27 \sqrt[3]{9 + x} - 54 \sqrt[3]{9 + x} + 27 \sqrt[3]{9 + x}$

Add $0$ and $x$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = x + 27 \sqrt[3]{9 + x} - 54 \sqrt[3]{9 + x} + 27 \sqrt[3]{9 + x}$

Subtract $54 \sqrt[3]{9 + x}$ from $27 \sqrt[3]{9 + x}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = x - 27 \sqrt[3]{9 + x} + 27 \sqrt[3]{9 + x}$

Combine the opposite terms in $x - 27 \sqrt[3]{9 + x} + 27 \sqrt[3]{9 + x}$ .

Add $- 27 \sqrt[3]{9 + x}$ and $27 \sqrt[3]{9 + x}$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\sqrt[3]{9 + x} + 3) = x$

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

Set up the composite result function.

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

Evaluate $f (x^{3} - 9 x^{2} + 27 x - 36)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{9 + x^{3} - 9 x^{2} + 27 x - 36} + 3$

Remove parentheses.

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{9 + x^{3} - 9 x^{2} + 27 x - 36} + 3$

Simplify each term.

Subtract $36$ from $9$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{x^{3} - 9 x^{2} + 27 x - 27} + 3$

Rewrite $x^{3} - 9 x^{2} + 27 x - 27$ in a factored form.

Factor $x^{3} - 9 x^{2} + 27 x - 27$ 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.

$p = \pm 1, \pm 27, \pm 3, \pm 9 q = \pm 1$

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

$\pm 1, \pm 27, \pm 3, \pm 9$

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

Substitute $3$ into the polynomial.

$3^{3} - 9 \cdot 3^{2} + 27 \cdot 3 - 27$

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

$27 - 9 \cdot 3^{2} + 27 \cdot 3 - 27$

Raise $3$ to the power of $2$ .

$27 - 9 \cdot 9 + 27 \cdot 3 - 27$

Multiply $- 9$ by $9$ .

$27 - 81 + 27 \cdot 3 - 27$

Subtract $81$ from $27$ .

$- 54 + 27 \cdot 3 - 27$

Multiply $27$ by $3$ .

$- 54 + 81 - 27$

Add $- 54$ and $81$ .

$27 - 27$

Subtract $27$ from $27$ .

$0$

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 9 x^{2} + 27 x - 27}{x - 3}$

Divide $x^{3} - 9 x^{2} + 27 x - 27$ by $x - 3$ .

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

$x$

$3$

$x^{3}$

$9 x^{2}$

$27 x$

$27$

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

					$x^{2}$
	$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			+	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					-	$6 x^{2}$	+	$18 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} + 18 x$

				$x^{2}$	-	$6 x$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$18 x$

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

				$x^{2}$	-	$6 x$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$18 x$
							+	$9 x$

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

				$x^{2}$	-	$6 x$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$18 x$
							+	$9 x$	-	$27$

Divide the highest order term in the dividend $9 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$	+	$9$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$18 x$
							+	$9 x$	-	$27$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$9$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$18 x$
							+	$9 x$	-	$27$
							+	$9 x$	-	$27$

The expression needs to be subtracted from the dividend, so change all the signs in $9 x - 27$

				$x^{2}$	-	$6 x$	+	$9$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$18 x$
							+	$9 x$	-	$27$
							-	$9 x$	+	$27$

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

				$x^{2}$	-	$6 x$	+	$9$
$x$	-	$3$		$x^{3}$	-	$9 x^{2}$	+	$27 x$	-	$27$
			-	$x^{3}$	+	$3 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$18 x$
							+	$9 x$	-	$27$
							-	$9 x$	+	$27$
										$0$

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

$x^{2} - 6 x + 9$

Write $x^{3} - 9 x^{2} + 27 x - 27$ as a set of factors.

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{(x - 3) (x^{2} - 6 x + 9)} + 3$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{(x - 3) (x^{2} - 6 x + 3^{2})} + 3$

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

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{(x - 3) (x^{2} - 2 \cdot x \cdot 3 + 3^{2})} + 3$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{(x - 3) {(x - 3)}^{2}} + 3$

Combine like factors.

Raise $x - 3$ to the power of $1$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{(x - 3) {(x - 3)}^{2}} + 3$

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

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{{(x - 3)}^{1 + 2}} + 3$

Add $1$ and $2$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = \sqrt[3]{{(x - 3)}^{3}} + 3$

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

$f (x^{3} - 9 x^{2} + 27 x - 36) = x - 3 + 3$

Combine the opposite terms in $x - 3 + 3$ .

Add $- 3$ and $3$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = x + 0$

Add $x$ and $0$ .

$f (x^{3} - 9 x^{2} + 27 x - 36) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{3} - 9 x^{2} + 27 x - 36$ is the inverse of $f (x) = \sqrt[3]{9 + x} + 3$ .

$f^{- 1} (x) = x^{3} - 9 x^{2} + 27 x - 36$

Find the Inverse f(x) = cube root of 9+x+3

Related Questions

Support LCMGCF.com by disabling your adblocker.