What is the enclosed volume of an irregular cube given the x,y,z coordinates of the 8 corners? I have the xyz coordinates of 8 points that forms an irregular-shaped cube.  This is an animation, so the cube is undergoing periodic or cyclical shape-change.   The co-planarism of any group or set of any combination of 4 of these points can't be assumed (it's a virtual certainty it practically never happens).
This question is very close to my situation:
Volume of irregular solid
but (a) has some very specific constraints, and (b) doesn't appear to have been answered despite showing very simple geometry.
Is there a simple formula for computing the internal volume from the xyz coordinates of the corners of this irregular cube?
 A: As pointed out by others in comment, if you don't place any constraint on the vertices/faces, there won't be a formula at all. 
When the shape is close to a cube,
one reasonable approximation has been suggested by vonbrand in the comment. 
Namely, linear interpolate between the 4 points delimiting a face. 
More precisely, given any quadlateral face $f$, let $\vec{A}_f$, $\vec{B}_f$, $\vec{C}_f$, $\vec{D}_f$ be the $4$ vertices of it, arranged in counterclockwise
orientation when viewed from outside of the "cube". We will approximate the face
by the one with following parametrization:
$$[0,1]\times[0,1] \ni (s,t) \quad\mapsto\quad \vec{X}(s,t) = (1-t)((1-s)\vec{A}_f + s\vec{B}_f) + t((1-s)\vec{D}_f + s\vec{C}_f)\in \mathbb{R}^3$$
Let $F$ be the collection of the six linear interpolated faces and $\Omega$ be the region bounded by them. As long as these faces
doesn't do anything strange like self-intersecting, the volume bounded by them is given by
$$\verb/Volume/ = \int_\Omega dV = \frac13 \int_\Omega \vec{\nabla}\cdot \vec{x} dV
= \frac13 \int_{\partial \Omega} \vec{x}\cdot d\vec{S}
= \frac13 \sum_{f\in F}\int_{[0,1]^2} \vec{X} \cdot \left(\partial_s \vec{X} \times \partial_t \vec{X}\right) ds dt
$$
If one struggle through the algebra, one find
$$\verb/Volume/ = \frac{1}{12}\sum_{f\in F} \bigg( [A_fB_fC_f] + [B_fC_fD_f] + [C_fD_fA_f] + [D_fA_fB_f]\bigg)\tag{*1}$$
where $[uvw]$ is the shorthand for the triple product $\vec{u}\cdot (\vec{v} \times \vec{w})$ of any three vectors $\vec{u},\vec{v},\vec{w}$.
Above formula has an interesting geometric interpretation. Given any face $ABCD$, there are two possible ways to triangular them, either as $\{ ABC, CDA \}$ or $\{ BCD, DAB \}$.
If one construct two tetrahedron with apex at origin and $ABC$, $CDA$ as base, the
sum of their volume equals to $\frac16 ( [ABC] + [CDA] )$. Similarly, if one construct two tetrahedron with apex at origin and $BCD$, $DAB$ as base, the sum of their volume equals to $\frac16 ( [BCD] + [DAB] )$. 
This means for each face $f$, one can interpret the expression
$$\frac{1}{12}\bigg( [A_fB_fC_f] + [B_fC_fD_f] + [C_fD_fA_f] + [D_fA_fB_f]\bigg)$$
as the "average" volume of the two possible ways of forming tetrahedron with apex
at origin and base at $A_fB_fC_fD_f$. The expression in $(*1)$ can be interpreted as the average of the volume under all possible triangulation of the faces.
