Find the Inverse x^3y=-9

$x^{3} y = - 9$

Divide each term by $x^{3}$ and simplify.

Tap for more steps…

Divide each term in $x^{3} y = - 9$ by $x^{3}$ .

$\frac{x^{3} y}{x^{3}} = \frac{- 9}{x^{3}}$

Cancel the common factor of $x^{3}$ .

Tap for more steps…

Cancel the common factor.

$\frac{x^{3} y}{x^{3}} = \frac{- 9}{x^{3}}$

Divide $y$ by $1$ .

$y = \frac{- 9}{x^{3}}$

Move the negative in front of the fraction.

$y = - \frac{9}{x^{3}}$

Interchange the variables.

$x = - \frac{9}{y^{3}}$

Solve for $y$ .

Tap for more steps…

Rewrite the equation as $- \frac{9}{y^{3}} = x$ .

$- \frac{9}{y^{3}} = x$

Solve for $y$ .

Tap for more steps…

Multiply each term by $y^{3}$ and simplify.

Tap for more steps…

Multiply each term in $- \frac{9}{y^{3}} = x$ by $y^{3}$ .

$- \frac{9}{y^{3}} \cdot y^{3} = x \cdot y^{3}$

Cancel the common factor of $y^{3}$ .

Tap for more steps…

Move the leading negative in $- \frac{9}{y^{3}}$ into the numerator.

$\frac{- 9}{y^{3}} \cdot y^{3} = x \cdot y^{3}$

Cancel the common factor.

$\frac{- 9}{y^{3}} \cdot y^{3} = x \cdot y^{3}$

Rewrite the expression.

$- 9 = x \cdot y^{3}$

$- 9 = x y^{3}$

Rewrite the equation as $x y^{3} = - 9$ .

$x y^{3} = - 9$

Divide each term by $x$ and simplify.

Tap for more steps…

Divide each term in $x y^{3} = - 9$ by $x$ .

$\frac{x y^{3}}{x} = \frac{- 9}{x}$

Cancel the common factor of $x$ .

Tap for more steps…

Cancel the common factor.

$\frac{x y^{3}}{x} = \frac{- 9}{x}$

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

$y^{3} = \frac{- 9}{x}$

Move the negative in front of the fraction.

$y^{3} = - \frac{9}{x}$

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

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

Simplify $\sqrt[3]{- \frac{9}{x}}$ .

Tap for more steps…

Rewrite $- \frac{9}{x}$ as ${({(- 1)}^{3})}^{3} \frac{9}{x}$ .

Tap for more steps…

Rewrite $- 1$ as ${(- 1)}^{3}$ .

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

Rewrite $- 1$ as ${(- 1)}^{3}$ .

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

Pull terms out from under the radical.

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

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

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

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

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

Multiply $\frac{\sqrt[3]{9}}{\sqrt[3]{x}}$ by $\frac{{\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{2}}$ .

$y = - (\frac{\sqrt[3]{9}}{\sqrt[3]{x}} \cdot \frac{{\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{2}})$

Combine and simplify the denominator.

Tap for more steps…

Multiply $\frac{\sqrt[3]{9}}{\sqrt[3]{x}}$ and $\frac{{\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{2}}$ .

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{\sqrt[3]{x} {\sqrt[3]{x}}^{2}}$

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

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{1} {\sqrt[3]{x}}^{2}}$

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

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{1 + 2}}$

Add $1$ and $2$ .

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{3}}$

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

Tap for more steps…

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

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{{(x^{\frac{1}{3}})}^{3}}$

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

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{x^{\frac{1}{3} \cdot 3}}$

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

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{x^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Tap for more steps…

Cancel the common factor.

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{x^{\frac{3}{3}}}$

Divide $1$ by $1$ .

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{x^{1}}$

Simplify.

$y = - \frac{\sqrt[3]{9} {\sqrt[3]{x}}^{2}}{x}$

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

$y = - \frac{\sqrt[3]{9} \sqrt[3]{x^{2}}}{x}$

Combine using the product rule for radicals.

$y = - \frac{\sqrt[3]{9 x^{2}}}{x}$

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

Tap for more steps…

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

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

Set up the composite result function.

$g (f (x))$

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

$g (- \frac{9}{x^{3}}) = - \frac{\sqrt[3]{9 {(- \frac{9}{x^{3}})}^{2}}}{- \frac{9}{x^{3}}}$

Multiply the numerator by the reciprocal of the denominator.

$g (- \frac{9}{x^{3}}) = - (\sqrt[3]{9 {(- \frac{9}{x^{3}})}^{2}} (- \frac{x^{3}}{9}))$

Rewrite using the commutative property of multiplication.

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 {(- \frac{9}{x^{3}})}^{2}} \frac{x^{3}}{9})$

Apply the product rule to $- \frac{9}{x^{3}}$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 {(- 1)}^{2} {(\frac{9}{x^{3}})}^{2}} \frac{x^{3}}{9})$

Simplify the expression.

Tap for more steps…

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

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 \cdot (1 {(\frac{9}{x^{3}})}^{2})} \frac{x^{3}}{9})$

Apply the product rule to $\frac{9}{x^{3}}$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 \cdot (1 (\frac{9^{2}}{{(x^{3})}^{2}}))} \frac{x^{3}}{9})$

Raise $9$ to the power of $2$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 \cdot (1 (\frac{81}{{(x^{3})}^{2}}))} \frac{x^{3}}{9})$

Multiply the exponents in ${(x^{3})}^{2}$ .

Tap for more steps…

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

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 \cdot (1 (\frac{81}{x^{3 \cdot 2}}))} \frac{x^{3}}{9})$

Multiply $3$ by $2$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 \cdot (1 (\frac{81}{x^{6}}))} \frac{x^{3}}{9})$

Multiply $9$ by $1$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{9 (\frac{81}{x^{6}})} \frac{x^{3}}{9})$

Combine $9$ and $\frac{81}{x^{6}}$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{\frac{9 \cdot 81}{x^{6}}} \frac{x^{3}}{9})$

Simplify the expression.

Tap for more steps…

Multiply $9$ by $81$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{\frac{729}{x^{6}}} \frac{x^{3}}{9})$

Rewrite $729$ as $9^{3}$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{\frac{9^{3}}{x^{6}}} \frac{x^{3}}{9})$

Rewrite $x^{6}$ as ${(x^{2})}^{3}$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{\frac{9^{3}}{{(x^{2})}^{3}}} \frac{x^{3}}{9})$

Rewrite $\frac{9^{3}}{{(x^{2})}^{3}}$ as ${(\frac{9}{x^{2}})}^{3}$ .

$g (- \frac{9}{x^{3}}) = - (- \sqrt[3]{{(\frac{9}{x^{2}})}^{3}} \frac{x^{3}}{9})$

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

$g (- \frac{9}{x^{3}}) = - (- \frac{9}{x^{2}} \cdot \frac{x^{3}}{9})$

Cancel the common factor of $9$ .

Tap for more steps…

Move the leading negative in $- \frac{9}{x^{2}}$ into the numerator.

$g (- \frac{9}{x^{3}}) = - (\frac{- 9}{x^{2}} \cdot \frac{x^{3}}{9})$

Factor $9$ out of $- 9$ .

$g (- \frac{9}{x^{3}}) = - (\frac{9 (- 1)}{x^{2}} \cdot \frac{x^{3}}{9})$

Cancel the common factor.

$g (- \frac{9}{x^{3}}) = - (\frac{9 \cdot - 1}{x^{2}} \cdot \frac{x^{3}}{9})$

Rewrite the expression.

$g (- \frac{9}{x^{3}}) = - (\frac{- 1}{x^{2}} \cdot x^{3})$

Cancel the common factor of $x^{2}$ .

Tap for more steps…

Factor $x^{2}$ out of $x^{3}$ .

$g (- \frac{9}{x^{3}}) = - (\frac{- 1}{x^{2}} \cdot (x^{2} x))$

Cancel the common factor.

$g (- \frac{9}{x^{3}}) = - (\frac{- 1}{x^{2}} \cdot (x^{2} x))$

Rewrite the expression.

$g (- \frac{9}{x^{3}}) = x$

Since $g (f (x)) = x$ , $f^{- 1} (x) = - \frac{\sqrt[3]{9 x^{2}}}{x}$ is the inverse of $f (x) = - \frac{9}{x^{3}}$ .

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

Find the Inverse x^3y=-9

Download our
App from the store

Create a High Performed UI/UX Design from a Silicon Valley.

Download our App from the store

Create a High Performed UI/UX Design from a Silicon Valley.

Download our
App from the store