Newton Raphson Works Great, Until It Doesn't - Part 2

Note that $x_{n}$ =0.2(-2)^{n}. Then the process not converge

Note that as $x$ goes from $0$ to $\pm\infty$ , the derivative of $\sqrt[3]{x}$ approaches $0$ monotonically. Therefore, after doing the Newton Raphson method one time from an arbitrary approximate root $x_0$ , we end up with the approximate root $x_1$ such that $\text{sgn}(x_1)=-\text{sgn}(x_0)$ and $|x_1| > |x_0|$ . This repeats to infinity, having $x_n$ become larger and larger while switching signs. Therefore, the method will not converge.

This solution is all just my intuition, not really any rigorous math, so apologies for any confusion.