Answer
\sqrt[8] 500 ≈ 2.17455928
Work Step by Step
First we need to formulate a function to use in Newton's method
x = \sqrt[8] 500
x^{8} - 500 = 0
Let f(x) = x^{8} - 500 = 0, The positive zero of this function is the solution to
x = \sqrt[8] 500
f'(x) = 8x^{7}
You could test some numbers or look at a graph to find a number to use as the initial approximation. Or just pick any number on interval [2,3] since it's a small interval. Since 2^{8} = 256 and 3^{8} = 6561
we can see that the number is closer to 2 than it is to 3.
We'll start with the first guess of x_{n} = x_{1} = 2.2 and plug into
x_{n+1} = x_{n} - (f(x_{n})/f'(x_{n})
until the numbers in the first 8 decimal places don't change anymore.
Each x_{n+1} we get will get plugged back into the formula to get x_{n+2}
x_{1} = 2.2
x_{2} = 2.2 - (f(x_{2.2})/f'(x_{2.2}) \approx 2.17556548
x_{3} = 2.17556548 - (f(x_{2.1755654874})/f'(x_{2.1755654874}) \approx 2.17456090
x_{4} = 2.17456090 - (f(x_{2.1745609034})/f'(x_{2.1745609034}) \approx 2.17455928
x_{5} = 2.17455928 - (f(x_{2.17455928})/f'(x_{2.17455928}) \approx
2.17455928
\sqrt[8] 500 = 2.17455928