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GATE CSE 1996 | Question: 2.5

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

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

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

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

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

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Answer is C. Direct application of the formula. With f(x)=X^{3}-c.
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