The disphenoid

A disphenoid is a non regular triangular pyramid whose faces (including the base) consist of congruent triangles.

The formula for the volume $V$ of disphenoid given its edge lengths $a$ , $b$ , and $c$ is:

$V\: = \: \sqrt{ \frac{ (a^2 \:+\: b^2 \:- \:c^2)(a^2\: -\: b^2 \:+ \:c^2)(-a^2\:+\:b^2\:+\:c^2)}{72}}$

However, this demands the values of the edge lengths to be known.

A more creative alternative to that would be using Heron's Formula, Newton's Sums, and this relationship

$16A^2R^2 \: = \: (abc)^2 \: + \: 9V^2$

where

A = area of each face

R = radius of the sphere circumscribing the disphenoid

V = volume of the disphenoid

To find $A$ , we use Heron's.

$A \: = \: \sqrt {s(s\:-\:a)(s\:-\:b)(s\:-\:c)}$

which again, demands the knowledge of the values of the side lengths. However, notice that these side lengths are the zeroes of the polynomial $P(x)$ . We could use that to our advantage and the formula becomes

$A \: = \: \sqrt{ s\cdot P(s) }$

where $s$ , by the way, is half the negative of the coefficient of $x^2$ . (Do you see why?)

For that we get $A = \sqrt{ \frac{ 2091}{16}}$ .

Now, the formula for the circumradius is

$R \: = \: \sqrt{\frac{a^2 \: + \:b^2 \: + \: c^2}{8}}$

which we can obtain using Newton's sums.

From that, we get,

$a^2 \: + \: b^2 \: + \: c^2 \: = \: 17(17) \: - \: 2(93)$

$a^2 \: + \: b^2 \: + \: c^2 \: =\: 103$

So, we get $R \: = \: \sqrt {\frac{103}{8}}$ .

Now, we're all set to find the volume!

$V^2 \: = \: \frac{16A^2 R^2 \: - \: (abc)^2 }{9}$

Substituting the values, we get

$V^2 \: = \: \frac{8005}{72}$

and $V =\large \boxed{10.544}$ .

Well, you can do this the "boring" way: determining the roots of the polynomial then proceeding to substitution. But I recommend otherwise!