Newton's Method: To solve the equation $$f(x) = 0$$ by Newton's Method we start with a good initial guess $x_0$ and then run the iteration $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \quad n=0,1,2,\ldots$$ until we get an approximation $x_{n+1}$ that is good enough for our purposes.
Suppose you want to compute the cube root of 4 by solving the equation $$x^3-4 = 0.$$ Since $1^3=1$ and $2^3=8$ Let's start with $$x_0 = 1.5$$ Then
$x_1 =$ ,
$x_2 =$ ,
$x_3 =$ , and
To check your answer compute $x_3^3 =$ .
Enter $x_1$, $x_2$ and $x_3$ with at least 6 correct digits beyond the decimal point.