Use the Newton-Raphson iteration method (NR) to find one positive root ( Xr > 0 ) of the equation $ X^3 +X -1 =0 $ ??
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Approach outline:
A plot (courtesy of Wolfram-Alpha) of the function shows:

Choose a starting point, say:
$$x_0 = 1$$
Write the Newton Iteration formula as:
$$x_{n+1} = x_n - \dfrac{f(x)}{f'(x)} = x_n - \dfrac{x_n^3 + x_n -1}{3 x_n^2 + 1}$$
Write out the iterates $x_0, x_1, x_2, \ldots$ and find:
$$x_5 = 0.682328$$
Amzoti
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Thanks for the right answer , i wish i had 15 reputation to give you vote up :( – Ahmed Osama Mar 17 '14 at 18:38
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Thank you so much , any way i am student faculty of mechanical engineering , thank you again 7 you are welcome too :D – Ahmed Osama Mar 17 '14 at 18:47
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Yes. Set $f(x)=x^3+x-1$ and iterate with the Newton Raphson formula:
$$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. $$ You need to guess a value for $x_0$ to start the iteration. Also, $f'(x_n)$ is the derivative of your function $f$ evaluated at $x_n$. If your iterations don't converge, try a different starting value $x_0$.
Hope this helps some.
Cheers,
Paul Safier
Paul Safier
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