Q2.5MCQMarks unknownNewton-Raphson iteration formula for finding c3\sqrt[3]{c}3c, where c>0c > 0c>0 is(A)xn+1=2xn3+c33xn2x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}xn+1=3xn22xn3+3c(B)xn+1=2xn3−c33xn2x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}xn+1=3xn22xn3−3c(C)xn+1=2xn3+c3xn2x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}xn+1=3xn22xn3+c(D)xn+1=2xn3−c3xn2x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}xn+1=3xn22xn3−cSign in to save attempts.Show answer← PreviousNext →SolutionsDiscussionLoading…Sign in with Google