`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.

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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.)

Notice that the function has two zeros. These two zeros are near x=-1.2 and x=0.8.

Let's solve for the approximate values of zeros of f(x) up to six decimal places.

For our first zero, let the initial value be -1.2.

`x_=-1.2`

`x_2= x_1- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=-1.221005982`

`x_3= x_2- (x_2^5+x_2^2-x_1)/(3x_2^4+1)=-1.220744126`

`x_4= x_3- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=-1.220744084`

Now there are two approximations that agree to six decimal places. So one of the approximate solution of the equation is **x=-1.220744** .

Next, let's solve for second zero. Let its initial value be 0.8.

`x_1=0.8`

`x_2= x_1- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=0.7247666905`

`x_3= x_2- (x_2^5+x_2^2-x_1)/(3x_2^4+1)=0.7244919491`

`x_4= x_3- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=0.7244919590`

There are already two approximations that have exact six decimal places. So, one of the solution is **x=0.724492** .

Therefore, the approximate solution to the equation `1/x=1+x^3` are `x={-1.220744,0.724492}` **.**