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`1/x = 1 + x^3` Use Newton's method to find all roots of the equation correct to six decimal places.

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`1/x=1+x^3`

Set the left side equal to zero.

`0=1+x^3 -1/x`

To solve this using Newton's method, apply the formula:

`x_(n+1) = x_n - (f(x_n))/(f'(x_n))`

Let the function be:

`f(x) = 1+x^3-1/x`

Take the derivative of f(x).

`f'(x) = 3x^2 +1/x^2`

Plug-in f(x) and f'(x) to the formula of Newton's method.

`x_(n+1)=x_n-(1+x_n^3-1/x_n)/(3x_n^2+1/x_n^2)`

This simplifies to:

`x_(n+1) = x_n- (x_n^5+x_n^2-x_n)/(3x_n^4+1)`

Then, refer to the graph of the function to get the initial values x when f(x) =0. (See attached figure.)

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