$\lfloor \sqrt[3]{|8x|}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200$ Find the $x$ :
$$\lfloor \sqrt[3]{|8x|}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200$$

$$x>0 , \to \lfloor \sqrt[3]{8x}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200 \\ \lfloor \sqrt[3]{8x}\rfloor \in \mathbb{Z}+\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor \in \mathbb{Z}=200$$
now what ?

We can write
$$
\left\{ \matrix{
  0 < x\quad  \Rightarrow \quad \left\lfloor {\root 3 \of {8x} } \right\rfloor  + \left\lfloor {\root 3 \of {\left\lfloor {8x} \right\rfloor } } \right\rfloor  = 200 \hfill \cr 
  x =  - y < 0\quad  \Rightarrow \quad \left\lfloor {\root 3 \of {8y} } \right\rfloor  + \left\lfloor {\root 3 \of {\left\lfloor { - 8y} \right\rfloor } } \right\rfloor  = 200 \hfill \cr}  \right.
$$
Now we have a fundamental property of the floor of a function which says
$$
\eqalign{
  & \left\{ \matrix{
  f(x){\rm  continuous}{\rm , monotone (strictly) increasing} \hfill \cr 
  f(x) = {\rm integer}\quad  \Rightarrow \quad x = {\rm integer} \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  \left\lfloor {f\left( {\left\lfloor x \right\rfloor } \right)} \right\rfloor  = \left\lfloor {f\left( x \right)} \right\rfloor  \hfill \cr 
  \left\lceil {f\left( {\left\lceil x \right\rceil } \right)} \right\rceil  = \left\lfloor {f\left( x \right)} \right\rfloor  \hfill \cr}  \right. \cr} 
$$
which can be easily proved considering that
$$
f\left( x \right) = \left\lfloor {f(x)} \right\rfloor  + \left\{ {f(x)} \right\} = f\left( {\left\lfloor x \right\rfloor  + \left\{ x \right\}} \right)
$$
That it is in particular the case with root function. So, for the case $0<x$
$$
\eqalign{
  & 200 = \left\lfloor {\root 3 \of {8x} } \right\rfloor  + \left\lfloor {\root 3 \of {\left\lfloor {8x} \right\rfloor } } \right\rfloor  = \left\lfloor {\root 3 \of {8x} } \right\rfloor  + \left\lfloor {\root 3 \of {8x} } \right\rfloor   \cr 
  & 100 = \left\lfloor {\root 3 \of {8x} } \right\rfloor   \cr 
  & 100 \le 2\root 3 \of x  < 101  \cr 
  & 50^{\,3}  \le x < \left( {50.5} \right)^{\,3}  \cr} 
$$
While the case $x<0$ does not provide any solution, since
$$
\eqalign{
  & 200 = \left\lfloor {\root 3 \of {8y} } \right\rfloor  + \left\lfloor {\root 3 \of {\left\lfloor { - 8y} \right\rfloor } } \right\rfloor  =   \cr 
  &  = \left\lfloor {\root 3 \of {8y} } \right\rfloor  + \left\lfloor {\root 3 \of { - \left\lceil {8y} \right\rceil } } \right\rfloor  =   \cr 
  &  = \left\lfloor {\root 3 \of {8y} } \right\rfloor  - \left\lceil {\root 3 \of {\left\lceil {8y} \right\rceil } } \right\rceil  =   \cr 
  &  = \left\lfloor {\root 3 \of {8y} } \right\rfloor  - \left\lceil {\root 3 \of {8y} } \right\rceil  \cr} 
$$
which would require
$$
 - 1 \le 200 = \left\lfloor {\root 3 \of {8y} } \right\rfloor  - \left\lceil {\root 3 \of {8y} } \right\rceil  \le 0
$$
