My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.

Proof (a). We have \[ \begin{align} g(\sqrt[3]{5}) &= \sqrt[3]{5} - c \frac{5 - 5}{5} = \sqrt[3]{5} - c \cdot 0 = \sqrt[3]{5}. \end{align} \] Thus, \(\sqrt[3]{5}\) is a fixed point of the function \( g(x) = x - c \frac{x^3 - 5}{x^2} \). \(\square\)

Solution (b). Let \(x_n = x\) and \(x_{n+1} = x - c \frac{x^3 - 5}{x^2}\). Since \(x^3 - 5 = (x - \sqrt[3]{5})(x^2 + \sqrt[3]{5}x + (\sqrt[3]{5})^2)\), we have Then, we have \[ \begin{align} \left|\dfrac{x - c \frac{x^3 - 5}{x^2} - \sqrt[3]{5}}{x - \sqrt[3]{5}}\right| = \left|1 - c \dfrac{x^2 + \sqrt[3]{5}x + (\sqrt[3]{5})^2}{x^2}\right| \end{align} \] In order to make the iteration converge, we need to have \(\lim\limits_{n \to \infty} \left|\dfrac{x_{n+1} - \sqrt[3]{5}}{x_n - \sqrt[3]{5}}\right| < 1\). Since it will converge when \(n\to \infty\), we have \(x\to \sqrt[3]{5}\). Thus, we have \[ \begin{align} \lim_{n \to \infty} \left|1 - c \dfrac{x^2 + \sqrt[3]{5}x + (\sqrt[3]{5})^2}{x^2}\right| < 1 \\ \lim_{x\to \sqrt[3]{5}} \left|1 - c \dfrac{x^2 + \sqrt[3]{5}x + (\sqrt[3]{5})^2}{x^2} \right| < 1 \\ \left|1 - c \dfrac{(\sqrt[3]{5})^2 + (\sqrt[3]{5})^2 + (\sqrt[3]{5})^2}{(\sqrt[3]{5})^2}\right| < 1 \\ \left|1 - c \dfrac{3(\sqrt[3]{5})^2}{(\sqrt[3]{5})^2}\right| = \left|1 - 3c \right| < 1 \\ -1 < 1 - 3c < 1 \\ 0< c < \frac{2}{3}. \end{align} \] Hence, in order to let the iteration converge, we need to have \(0 < c < \frac{2}{3}\).

Solution (c).