<s>[INST] Volume of a tetrahedron whose 4 faces are congruent. Suppose that I have a tetrahedron such that all four faces consist of congruent triangles, says with the lengths $a,b$ and $c$ for each side. Is there a beautiful method to compute its volume? 
PS. The reason for me tagging calculus and linear algebra is that I figured that the technique used to calculate such a problem may come from this areas. 

 [/INST] If the given triangle is acute-angled, define positive numbers $x, y, z$ by:
\begin{align*}
x^2 & = \tfrac{1}{2}(b^2 + c^2 - a^2), \\
y^2 & = \tfrac{1}{2}(c^2 + a^2 - b^2), \\
z^2 & = \tfrac{1}{2}(a^2 + b^2 - c^2).
\end{align*}
Let $\mathbf{u}, \mathbf{v}, \mathbf{w}$ be mutually orthogonal vectors of unit length, and define:
$$
\mathbf{t}_{ijk} = ix\mathbf{u} + jy\mathbf{v} + kz\mathbf{w} \quad (i, j, k = 0, 1).
$$
The cuboid with these 8 vertices has volume $xyz$. It contains a tetrahedron $T$ with vertices $\mathbf{t}_{100}\mathbf{t}_{010}\mathbf{t}_{001}\mathbf{t}_{111}$, and faces which are triangles with the same sides:
\begin{align*}
a & = \sqrt{y^2 + z^2}, \\
b & = \sqrt{z^2 + x^2}, \\
c & = \sqrt{x^2 + y^2}.
\end{align*}
The remainder of the cuboid consists of 4 tetrahedra, each with sides $x, y, z$ meeting at right angles at one of the vertices $\mathbf{t}_{000}, \mathbf{t}_{011}, \mathbf{t}_{101}, \mathbf{t}_{110}$, and each therefore having volume $\tfrac{1}{6}xyz$, implying that $T$ has volume $\tfrac{1}{3}xyz$.
Wikipedia illustration (with different notation):


Another Wikipedia illustration, to accompany the comment:

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