Is $\left||x-y|+|x+y|-2z\right|+\left||x-y|+|x+y|+2z\right|=1$ the equation of a cube? Is this the equation for a unit cube of edge length $2$?
$$\left|\,\left|x-y\right|+\left|x+y\right|-2z\,\right|\;+\;\left|\,\left|x-y\right|+\left|x+y\right|+2z\,\right|\;=\;1$$
Online tools unable to show the shape of a cube.
 A: Method 1: Consider the 2-D version first. Show that $|x-y|+|x+y|=1$ gives a square, but without resorting to graphing it.    
Hint: Prove that $|x+y | + |x-y| = 2 \max(|x|,|y|)$
Extend this to 3 dimensions.     

 Find an equation that represents $ \max(|x|,|y|,|z|) = k$.    

Bonus: What is the extension to 4 dimensions? 

Method 2: As Saulspatz suggested, guess the vertices of the cube, and verify that a point on any of the 12 edges satisfies the conditions of the problem. 
A: An equation of the form $|x+y|+|x-y|=a$ describes an axis-aligned square, as can be shown by the change of variable $u:=x+y,v=x-y$, which is a similarity. (By symmetry, $|u|+|v|=a$ is a square diamond.)
Now if $2s:=|x-y|+|x+y|\ge0$, the locus of $|2s+2z|+|2s-2z|=1$ is a half axis-aligned square, and any horizontal section is a square.
A: Yep... but of edge length $1/2$, not $2$.  In Mathematica:
RegionPlot3D[
   Abs[Abs[x - y] + Abs[x + y] - 2 z] + 
   Abs[Abs[x - y] + Abs[x + y] + 2 z] < 1,
 {x, -.3, .3}, {y, -.3, .3}, {z, -.3, .3},
 PlotPoints -> 100,
 PlotStyle -> Opacity[0.5]]


