Solve for $~f(f(100))~$ when the function is given as $$f(x)= ([\{x\}] + 3 - x^7)^{1/7}~.$$
I know that $~[\{x\}]~$ will result in $0$, but I am not able to solve further. Also the answer is extremely simplified and is probably $100$ or some other whole number. So complex answers which require a calculator to be solved won’t work. This problem doesn’t require a calculator, and can be compared to simple identities (eg. $\sin^2x + \cos^2x = 1$).
 A: Thanks to Luca Bresson for pointing out the obvious.
The question as you asked it can't be correct with the story you have told in the comments.
I imagine you had a typo and the question was:
$f(x) = ([\{x\}] + 3 - x^7)^{\frac 17}$.
If so....
As you note $[\{x\}] = 0$ always.
So $f(100) = ([\{100\}] + 3 - 100^7)^{\frac 17} = \color{blue}{(3-100^7)^{\frac 17}}$.
Ad so $f(f(100)) =([\{f(100)\}] + 3 - f(100)^7)^{\frac 17}=$
$(3 - (\color{blue}{(3-100^7)^{\frac 17}})^7)^{\frac 17} =$
$(3 - \color{blue}{(3-100^7)}^1)^{\frac 17} = $
$(3 - \color{blue}{(3-100^7)})^{\frac 17} =$
$(\color{blue}{100^7})^{\frac 17} = \color{blue}{100}^1 = \color{blue}{100}$.
Indeed it will always be the case that 
$f(f(K)) = K$.
$f(f(K)) = (3 - f(K)^7)^{\frac 17} =$
$(3 - ((3 - K^7)^{\frac 17})^7)^{\frac 17} =$
$(3 - (3 -K^7))^{\frac 17} =$
$(K^7)^{\frac 17} = K$.
......
Furthermore we use the notation $f^{\underline m}(K)$ to mean $\underbrace {f(f(f(.....(f}_{m\text{ times}}(K))....)))$ then we will have $f^{\underline{2k}}(K) = K$ and $f^{\underline{2k + 1}}(K) = f(K)$.  $f(K)= (3 - K^7)^{\frac 17}$ can be nasty and if $K$ is a positive whole number it can be a negative number with lots of decimals.  But iterating it an even number of times will always "undo" it.
A: $$f(100)=([\{100\}] + 3 -10000)^{\frac17}=(-9997)^{\frac17}\approx-3.72743 $$
$$f(f(100))=([\{-3.72743\}]+3-(-3.72743)^2)^{\frac17}=([0.27257]+3-13.89276529)^{\frac17}\approx-1.4065740$$
