In the inverse we replace the place of x by y .
[tex]g(x) = \sqrt[3]{x} - 3 \\ \\ x = \sqrt[3]{y} - 3 \\ \sqrt[3]{y} = x + 3 \\ y = {(x + 3)}^{3} \\ y = {(x + 3)}^{2} (x + 3) \\ y = ({x}^{2} + 6x + 9)(x + 3) \\ \\ y = {x}^{3} + 6 {x}^{2} + 9x + 3 {x}^{2} + 18x + 27 \\ \\ y = {x}^{3} + 9 {x}^{2} + 27x + 27 \\ \\ \\ g(x)^{ - 1} = {x}^{3} + 9 {x}^{2} + 27x + 27 [/tex]
I hope I helped you^_^
