<s>[INST] Volume of a solid created by an extended tetrahedron 
Every edge of a tetrahedron with length $p$ is extended through the vertices by $p$.
Now all 12 points create a new solid $J$ of which I seek the volume dependent on the volume of the tetrahedron in the centre.


With some help the solution becomes clear:
The whole Volume of all pyramids is:
$$V_P=4\cdot\frac{1}{6}\cdot\left(\frac{a}{\sqrt{2}}\right)^3 +4\cdot\frac{1}{6}\cdot\left(\sqrt{2}a\right)^3 =\frac{3}{\sqrt{2}}a^3 \tag{1}$$
The side of the large cube is
$$s=\frac{3}{\sqrt{2}}a\tag{2}$$
and the Volume is respectively
$$V_C=\left(\frac{3}{\sqrt{2}}a\right)^3=\frac{27\sqrt{2}}{4}\cdot a^3 \tag{3}$$
The last step is subtracting the Volume of the pyramids.
$$V_J=V_C-V_P= \frac{27\sqrt{2}a^3}{4}-\frac{3}{\sqrt{2}}a^3=\frac{21\sqrt{2}a^3}{4} \tag{4}$$
Thx for the help.
 [/INST] There are only three types of faces for the new solid.  They are either equilateral triangles of side lenghth $p$ that are $\dfrac{7}{4} h$ away from the center, where $h = p \sqrt{\dfrac{2}{3}} $, or equilateral triangles of side $2p$ that are $ \frac{5}{4} h $ from the centroid, or rectangles of sides $ p $ and $ 2 p $ that are $ \dfrac{3 \sqrt{2}}{4 } p $ from the center.
$V = (1/3) \left(4 (\dfrac{\sqrt{3}}{4}) ( p^2 h (\dfrac{7}{4}) + 5 p^2 h ) + 6 (2 p^2)( \dfrac{3\sqrt{2}}{4} p ) \right) = \sqrt{3} ( \dfrac{9}{4} p^2 h ) + 3 \sqrt{2} p^3 $
$V =  \sqrt{2} p^3 ( \dfrac{21}{4} )   = 63 V_0$
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