SOLUTION 3: If we let $x=100^{1/5}$, then $x^5=100$ and $x^5-100=0$, so let's define function $f(x) = x^5-100$, whose graph is given below.

The derivative of $f$ is $f'(x) = 5x^4$. Now use Newton's Method: $$x_{n+1} = x_{n} - { f(x_{n}) \over f'(x_{n}) } \ \ \ \ \longrightarrow$$ $$x_{n+1} = x_{n} - { x_{n}^5-100 \over 5x_{n}^4 } \ \ \ \ \longrightarrow$$ (Let's simplify the right-hand side of this equation. First get a common denominator.) $$x_{n+1} = x_{n} \ { 5x_{n}^4 \over 5x_{n}^4 } - { x_{n}^5-100 \over 5x_{n}^4 } \ \ \ \ \longrightarrow$$ $$x_{n+1} = { 5x_{n}^5 - ( x_{n}^5-100 ) \over 5x_{n}^4 } \ \ \ \ \longrightarrow$$ $$x_{n+1} = { 4x_{n}^5 + 100 \over 5x_{n}^4 }$$ I will choose to let $x_{0}=2$. Using Newton's Method formula for 6 iterations in a spreadsheet results in :

Thus $\ 100^{1/5} \$ to eight decimal places is $\ 100^{1/5} \approx 2.511886432$.