Answer: [tex]f^{-1}(x) = \frac{\sqrt[3]{x^{2}}}{x}[/tex]
Step-by-step explanation:
[tex]y = \frac{1}{x^{3}}[/tex]
Inverse is when you swap the x's and y's and then solve for "y":
[tex]x = \frac{1}{y^{3}}[/tex]
[tex]y^{3} = \frac{1}{x}[/tex]
[tex]y = \frac{1}{\sqrt[3]{x}}[/tex]
[tex]y = \frac{1}{\sqrt[3]{x}}*(\frac{\sqrt[3]{x}}{\sqrt[3]{x}})^{2}[/tex]
[tex]y = \frac{\sqrt[3]{x^{2}}}{x}[/tex]
