Extended Pythagorean Theorem Extended Pythagorean Theorem We all well familiar with the basic Pytha

As achille hui pointed out in a comment, the relevant generalization here would be De Gua's theorem for three dimensions or the Conant–Beyer theorem for arbitrary dimensions. Colloquially speaking it states that the squared measure of your object equals the sum of the squared measures of its projection on a set of mutually orthogonal hyperplanes, in your case the coordinate axes.
So let's look at this in the dimensions we know. A segment from (a,0) to (0,b) has some length d as measure. Projecting the segment onto the x axis yields a segment of length a, projecting onto the y axis yields a segment of length b. So taken together you have $d^2=a^2+b^2$ as expected.
In three dimensions, you have a triangle, and project that onto the coordinate planes. But here the area of the projection is a triangle, not a rectangle, so you don't simply multiply the intercepts, but divide by 2 as the area of a right triangle is $\frac{1}{2}ab$. Thus the area is
$s^2=(\frac{ab}{2}{)}^2+(\frac{ac}{2}{)}^2+(\frac{bc}{2}{)}^2=\frac{a^2{b}^2+a^2{c}^2+b^2{c}^2}{4}$
This differs from the $s^2=\frac{1}{4}(a^2+b^2+c^2)$ you claimed here originally, but fits in with the $s^2=\frac{1}{4}((ab{)}^2+(bc{)}^2+(ac{)}^2)$ you wrote in the duplicate post.
For four dimensions, you have to turn the volume of the cuboid into the volume of the simplex spanned by three of its edges. The factor to divide by here is 6, so the formula for the volume will be
$v^2=(\frac{abc}{6}{)}^2+(\frac{abd}{6}{)}^2+(\frac{acd}{6}{)}^2+(\frac{bcd}{6}{)}^2$
So where you assumed the denominator to be 9, it is in fact $6^2=36$. That 6 there in the denominator is the 3! achille hui wrote in the comment, which is a more useful notation because it also explains how you get the corresponding factor for arbitrary dimensions. You might think of the area of the simplex as $\frac{1}{3}$ times base plane times hright, where the base plane is again $\frac{1}{2}$ times one length times the other. If you go to d dimensions, you take all ways of combining d−1 of the d edge lengths, multiply them and divide them by (d−1)!. The resulting numbers get squared and added. If you want a general formula for arbitrary dimensions, you could write something like
$v^2=\underset{i=1}{\overset{d}{\sum <!--\; \sum \; -->}}{\left(\begin{array}{c}{\displaystyle \underset{\begin{array}{c}j=1\\ j\ne <!--\; \ne \; -->i\end{array}}{\overset{d}{\prod <!--\; \prod \; -->}}{x}_j}\\ (d-<!--\; -\; -->1)!\end{array}\right)}^2$
This is for the simplex spanned by $(x_1,0,\dots <!--\; \dots \; -->,0)$ through $(0,\dots <!--\; \dots \; -->,0,x_d)$.