Calculus 8th Edition

Let $x_1\neq x_2$. Then $g(x_1)=\sqrt[3]{x_1}$ and $g(x_2)=\sqrt[3]{x_2}$. Exponentianting and then subtracting we have: $$g(x_1)^3-g(x_2)^3=x_1-x_2\neq 0,\text{ because }x_1\neq x_2.$$ For $g(x_1)^3-g(x_2)^3\neq 0$ to hold, the only possbility is that $g(x_1)\neq g(x_2)$ (because we have them exponentiated to the odd power), so different arguments always map to different values of the function meaning that $g$ is one-to-one.