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 $.
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