Let $T$ be a tetrahedron with sidelength $s$. It's volume is proportional to $s^3$. $$V(T)=ks^3$$ Let $T_1$ and $T_2$ be tetrahedra with sidelength $s_1=\frac 13 s$ and $s_2=\frac 23 s$
Can't we decompose $V(T)$ as follows?
$$V(T)=ks^3=4V(T_2)-6V(T_1)=4k(\frac 23s)^3-6k(\frac 13s)^3$$
The idea behind that is to count 4 $T_2$ inside of $T$ and subtract the overlaps $T_1$ that happen at the edges of $T$.
1 shows an example of $T$ with sidelength $s=\sqrt{2}$
2 shows $T_{2A}=T_2$ which shares the corner $A$ with $T$.
3 shows $T_{2B}=T_2$ which shares the corner $B$ with $T$.
Each of the 4 large $T_2$ shares 1 corner with $T$.
Every 2 instances of $T_2$ overlap in a small region $T_1$ like so
4 shows $T_{1AB}$ which shares the edge $AB$ with $T$.
I must be missing something however since the equation above yields $\frac{26}{27}ks^3$
What am I missing?
Edit: Found it! :) Visualizing a bit better in geogebra, as @ChrisLewis suggested, showed it is the topological dual of $T$ hidden in the center which is also a tetrahedron $T_1$. It's volume is of course exactly what was missing before.
5 shows that 3 $T_2$ form a valley which the 4th one with it's flat bottom surface can't fill completly. By the way this region is of course also where the pivotmechanism in the Pyraminx is hidden.
So we got: $$V(T)=ks^3=4V(T_2)-6V(T_1)+V(T_1)=4k(\frac 23s)^3-6k(\frac 13s)^3+k(\frac 13s)^3=ks^3$$
I think it goes like $$4×8-6×1+4×0-1×0+1=27.$$
I imagined Pyraminx. I think, the piece in the middle does not belong to any T2. Right?
By the way, I am bad at solving Rubik's cube.