Suppose you didn’t know that 3√64 = 4. Write down Newton’s method for solving f (x) =x3 − 64 = 0. Choose x0 = 5 as your initial guess for the solution of this equation and iterateuntil the first 5 significant figures of your solution are correct.

Newton's method is a root-finding algorithm which produces successively better approximations to the roots (or zeroes) of a real-valued function. The method starts with a function f defined over the real numbers x, an initial guess x0 for a root of f, and the derivative f' of f.

The function f we're trying to find the root for is f(x) = x^3 - 64. The derivative of this function, f'(x), is 3x^2.

Newton's method formula is: x_n+1 = x_n - f(x_n) / f'(x_n)

Let's start with x0 = 5.

1. First iteration (n=0): x1 = x0 - f(x0) / f'(x0) = 5 - (5^3 - 64) / (3*5^2) = 5 - (125 - 64) / 75 = 5 - 61 / 75 = 5 - 0.813333333 = 4.186666667

2. Second iteration (n=1): x2 = x1 - f(x1) / f'(x1) = 4.186666667 - ((4.186666667)^3 - 64) / (3*(4.186666667)^2) = 4.186666667 - (73.17911111 - 64) / 52.49377778 = 4.186666667 - 9.17911111 / 52.49377778 = 4.186666667 - 0.174931111 = 4.011735556

3. Third iteration (n=2): x3 = x2 - f(x2) / f'(x2) = 4.011735556 - ((4.011735556)^3 - 64) / (3*(4.011735556)^2) = 4.011735556 - (64.56411111 - 64) / 48.14111111 = 4.011735556 - 0.56411111 / 48.14111111 = 4.011735556 - 0.011711111 = 4.000024444

4. Fourth iteration (n=3): x4 = x3 - f(x3) / f'(x3) = 4.000024444 - ((4.000024444)^3 - 64) / (3*(4.000024444)^2) = 4.000024444 - (64.00029496 - 64) / 48.00029496 = 4.000024444 - 0.00029496 / 48.00029496 = 4.000024444 - 0.000006144 = 4.0000183

5. Fifth iteration (n=4): x5 = x4 - f(x4) / f'(x4) = 4.0000183 - ((4.0000183)^3 - 64) / (3*(4.0000183)^2) = 4.0000183 - (64.0000002 - 64) / 48.0000002 = 4.0000183 - 0.0000002 / 48.0000002 = 4.0000183 - 0.00000000416 = 4.000018296

After 5 iterations, the first 5 significant figures are correct (4.0000).