Find the Inverse x^3y=-4

$x^{3} y = - 4$

Divide each term in $x^{3} y = - 4$ by $x^{3}$ and simplify.

Divide each term in $x^{3} y = - 4$ by $x^{3}$ .

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as $- \frac{4}{y^{3}} = x$ .

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

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y^{3}, 1$

The LCM of one and any expression is the expression.

$y^{3}$

Multiply each term in $- \frac{4}{y^{3}} = x$ by $y^{3}$ to eliminate the fractions.

Multiply each term in $- \frac{4}{y^{3}} = x$ by $y^{3}$ .

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

Simplify the left side.

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

Move the leading negative in $- \frac{4}{y^{3}}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$- 4 = x y^{3}$

Solve the equation.

Rewrite the equation as $x y^{3} = - 4$ .

$x y^{3} = - 4$

Divide each term in $x y^{3} = - 4$ by $x$ and simplify.

Divide each term in $x y^{3} = - 4$ by $x$ .

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

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

Pull terms out from under the radical.

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

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

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

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

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

Multiply $\frac{\sqrt[3]{4}}{\sqrt[3]{x}}$ by $\frac{{\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{2}}$ .

$y = - (\frac{\sqrt[3]{4}}{\sqrt[3]{x}} \cdot \frac{{\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{2}})$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{4}}{\sqrt[3]{x}}$ by $\frac{{\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{2}}$ .

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{\sqrt[3]{x} {\sqrt[3]{x}}^{2}}$

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

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{1} {\sqrt[3]{x}}^{2}}$

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

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{1 + 2}}$

Add $1$ and $2$ .

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{{\sqrt[3]{x}}^{3}}$

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

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

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{{(x^{\frac{1}{3}})}^{3}}$

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

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{x^{\frac{1}{3} \cdot 3}}$

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

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{x^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{x^{\frac{3}{3}}}$

Rewrite the expression.

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{x^{1}}$

Simplify.

$y = - \frac{\sqrt[3]{4} {\sqrt[3]{x}}^{2}}{x}$

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

$y = - \frac{\sqrt[3]{4} \sqrt[3]{x^{2}}}{x}$

Combine using the product rule for radicals.

$y = - \frac{\sqrt[3]{4 x^{2}}}{x}$

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

$f^{- 1} (x) = - \frac{\sqrt[3]{4 x^{2}}}{x}$

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

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

$f^{- 1} (- \frac{4}{x^{3}}) = - \frac{\sqrt[3]{4 {(- \frac{4}{x^{3}})}^{2}}}{- \frac{4}{x^{3}}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (- \frac{4}{x^{3}}) = - (\sqrt[3]{4 {(- \frac{4}{x^{3}})}^{2}} (- \frac{x^{3}}{4}))$

Rewrite using the commutative property of multiplication.

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{4 {(- \frac{4}{x^{3}})}^{2}} \frac{x^{3}}{4})$

Apply the product rule to $- \frac{4}{x^{3}}$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{4 {(- 1)}^{2} {(\frac{4}{x^{3}})}^{2}} \frac{x^{3}}{4})$

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

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

Apply the product rule to $\frac{4}{x^{3}}$ .

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

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

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

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

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

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

Multiply $3$ by $2$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{4 \cdot (1 (\frac{16}{x^{6}}))} \frac{x^{3}}{4})$

Combine exponents.

Multiply $4$ by $1$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{4 (\frac{16}{x^{6}})} \frac{x^{3}}{4})$

Combine $4$ and $\frac{16}{x^{6}}$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{\frac{4 \cdot 16}{x^{6}}} \frac{x^{3}}{4})$

Multiply $4$ by $16$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{\frac{64}{x^{6}}} \frac{x^{3}}{4})$

Rewrite $64$ as $4^{3}$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{\frac{4^{3}}{x^{6}}} \frac{x^{3}}{4})$

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

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{\frac{4^{3}}{{(x^{2})}^{3}}} \frac{x^{3}}{4})$

Rewrite $\frac{4^{3}}{{(x^{2})}^{3}}$ as ${(\frac{4}{x^{2}})}^{3}$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \sqrt[3]{{(\frac{4}{x^{2}})}^{3}} \frac{x^{3}}{4})$

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

$f^{- 1} (- \frac{4}{x^{3}}) = - (- \frac{4}{x^{2}} \cdot \frac{x^{3}}{4})$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{4}{x^{2}}$ into the numerator.

$f^{- 1} (- \frac{4}{x^{3}}) = - (\frac{- 4}{x^{2}} \cdot \frac{x^{3}}{4})$

Factor $4$ out of $- 4$ .

$f^{- 1} (- \frac{4}{x^{3}}) = - (\frac{4 (- 1)}{x^{2}} \cdot \frac{x^{3}}{4})$

Cancel the common factor.

$f^{- 1} (- \frac{4}{x^{3}}) = - (\frac{4 \cdot - 1}{x^{2}} \cdot \frac{x^{3}}{4})$

Rewrite the expression.

$f^{- 1} (- \frac{4}{x^{3}}) = - (\frac{- 1}{x^{2}} \cdot x^{3})$

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

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

$f^{- 1} (- \frac{4}{x^{3}}) = - (\frac{- 1}{x^{2}} \cdot (x^{2} x))$

Cancel the common factor.

$f^{- 1} (- \frac{4}{x^{3}}) = - (\frac{- 1}{x^{2}} \cdot (x^{2} x))$

Rewrite the expression.

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

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

Set up the composite result function.

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

Evaluate $f (- \frac{\sqrt[3]{4 x^{2}}}{x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{{(- \frac{\sqrt[3]{4 x^{2}}}{x})}^{3}}$

Simplify the denominator.

Apply the product rule to $- \frac{\sqrt[3]{4 x^{2}}}{x}$ .

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{{(- 1)}^{3} {(\frac{\sqrt[3]{4 x^{2}}}{x})}^{3}}$

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

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- {(\frac{\sqrt[3]{4 x^{2}}}{x})}^{3}}$

Apply the product rule to $\frac{\sqrt[3]{4 x^{2}}}{x}$ .

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{{\sqrt[3]{4 x^{2}}}^{3}}{x^{3}}}$

Rewrite ${\sqrt[3]{4 x^{2}}}^{3}$ as $4 x^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 x^{2}}$ as ${(4 x^{2})}^{\frac{1}{3}}$ .

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{{({(4 x^{2})}^{\frac{1}{3}})}^{3}}{x^{3}}}$

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

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

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

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{{(4 x^{2})}^{\frac{3}{3}}}{x^{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{{(4 x^{2})}^{\frac{3}{3}}}{x^{3}}}$

Rewrite the expression.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{4 x^{2}}{x^{3}}}$

Simplify.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{4 x^{2}}{x^{3}}}$

Cancel the common factor of $x^{2}$ and $x^{3}$ .

Factor $x^{2}$ out of $4 x^{2}$ .

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{x^{2} \cdot 4}{x^{3}}}$

Cancel the common factors.

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

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{x^{2} \cdot 4}{x^{2} x}}$

Cancel the common factor.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{x^{2} \cdot 4}{x^{2} x}}$

Rewrite the expression.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - \frac{4}{- \frac{4}{x}}$

Multiply the numerator by the reciprocal of the denominator.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - (4 (- \frac{x}{4}))$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{x}{4}$ into the numerator.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - (4 (\frac{- x}{4}))$

Cancel the common factor.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = - (4 (\frac{- x}{4}))$

Rewrite the expression.

$f (- \frac{\sqrt[3]{4 x^{2}}}{x}) = x$

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

$f^{- 1} (x) = - \frac{\sqrt[3]{4 x^{2}}}{x}$

Find the Inverse x^3y=-4

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