Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is

$x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$

$x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$

$x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$

$x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$

Answer: C