$\DeclareMathOperator\Conv{Conv}$I am finding all 3-dimensional symmetric reflexive polytopes. To do so, first, we know that all 2 dim symmetric reflexive polytopes are $X_3=\Conv((-1,-1),(1,0),(0,1))$, $X_4=\Conv((0, \pm 1), (\pm 1, 0))$, $X_6=\Conv ((\pm 1,0), (0, \pm 1),(1,-1), (-1,1))$, $X_8=[-1, 1]\times [-1,1]$ and $X_9=\Conv((-1,-1), (2,-1), (-1,2)$.
Now we can produce 3 dimensional symmetric reflexive polytopes by $X \times [-1,1]$, or the dual which is $D(X):=\Conv ( [ (p,0)| p\in X ], (0,0, \pm 1) $ (I don't know why I cannot type the set symbol). we now count the remaining term:
$P^3= \Conv((-1,-1,-1), (3,-1,-1), (-1,3,-1), (-1,-1,3)$, $P^3/ \mathbb{Z}_4=\Conv((-1,-1,-1), (1,0,0), (0,1,0),(0,0,1))$ (or the dual of $P^3$).
The remaining is those I start not quite sure: first, we can blow up 4 points of $P^3$ which I can still understand (Cut 4 simplex on each vertex), (and I call it $B_1$) but what is the dual of this?
Also, seem $[-1,1]\times [-1,1] \times [-1,1]$ blowup 2 points (cut the [-1,-1,-1] and [1,1,1]) still be symmetric? and what is the dual? Also I expect I can blow up $B_1$ again to get something, and its dual is different?
Third I expect there is a self dual which is a analog of $X_6$, and it should have 24 simplex touch the 0, and it is the intersection of all space $ v \cdot (x,y,z) \leq 1$ and $w \dots (x,y,z) \geq -1$, where $v=v_1, v_2,v_3$, $v_i=0,1$, but I am not sure if I count it before.
Finally are there any polytopes more I miss? Thank you.
