4 votes 4 votes Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is $x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$ Numerical Methods gate1996 numerical-methods newton-raphson normal out-of-syllabus-now + – Kathleen asked Oct 9, 2014 • retagged Dec 9, 2015 by amarVashishth Kathleen 1.8k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes Answer: C Rajarshi Sarkar answered Jun 3, 2015 Rajarshi Sarkar comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Answer is C. Direct application of the formula. With f(x)=X^{3}-c. kireeti answered Oct 26, 2014 kireeti comment Share Follow See all 0 reply Please log in or register to add a comment.