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I am wondering how to evaluate this integral. Wolfram Alpha says it is $fracpi^3/22$ but I have no idea how get there.

$$\
int_0^infty int_0^infty int_0^infty e^-(xy+yz+zx) dx dy dz\
$$

I guess it may have some connection with
$$int_0^infty int_0^infty int_0^infty e^-(x^2+y^2+z^2) dx dy dz=left(int_0^infty e^-x^2 dxright)^3=fracpi^3/28$$
and/or
$$int_0^infty int_0^infty int_0^infty e^-(x+y+z)^2 dx dy dz=dots=fracsqrtpi8$$
but I am not sure. Thanks in advance.

Taking @ersh's suggestion, I did:
beginalign
&int_0^infty int_0^infty int_0^infty e^-(xy+yz+zx) dx dy dz\
&=int_0^infty int_0^infty e^-yzint_0^infty e^-(y+z)x dx dy dz\
&=int_0^infty int_0^infty frace^-yzy+z dy dz=int_0^infty int_z^infty frace^-(y-z)zy dy dz\
&=int_0^infty e^z^2int_z^infty frace^-yzy dy dz\
endalign
Now let me work on this integral:
beginalign
I(a)&=int_z^infty frace^-ayzy dy\
fracdI(a)da&=int_z^infty ze^-ayz dy=left[-frace^-ayzaright]_z^infty=frace^-az^2a\
lim_ato0I(a)&=-z^2\
therefore I(a)&=
endalign
Wait. I'm in the loop!

Here's another approach: By making substitution $$
(u,v,w) = (xy,yz,zx),
$$ we have$$mathrm du mathrm dvmathrm dw = 2xyz mathrm dxmathrm dymathrm dzimplies fracmathrm du mathrm dvmathrm dw2sqrtuvw=mathrm dxmathrm dymathrm dz.$$ Hence,
$$beginalign*
int_0^infty int_0^infty int_0^infty e^-(xy+yz+zx)mathrm dxmathrm dymathrm dz&=frac12int_0^infty int_0^infty int_0^infty frace^-u-v-wmathrm du mathrm dvmathrm dwsqrtuvw\&=frac12left( int_0^infty frace^-umathrm dusqrturight)^3\&=frac12left(2 int_0^infty e^-v^2mathrm dvright)^3\
&=fracpi^frac322.
endalign*$$

A chain of mostly elementary manipulations:
beginalign*I &= int_0^inftyint_0^inftyint_0^inftye^-(xy+yz+zx),dx,dy,dz\
&= int_0^inftyint_0^inftyint_0^inftye^-yze^-x(y+z),dx,dy,dz\
&= int_0^inftyint_0^inftyfrace^-yzy+z,dy,dzendalign*
First, we evaluate the inner integral with elementary techniques. There's some clutter that amounts to constant multipliers, but it's just $e^ax$ at its core. Next, to handle $y$, we substitute $y=zu$, $dy=z,du$:
beginalign*I &= int_0^inftyint_0^infty frace^-z^2uz(1+u)cdot z,du,dz\
&= int_0^inftyint_0^infty frace^-uz^21+u,dz,du\
&= int_0^inftyfracfracsqrtpi2sqrtu1+u,duendalign*
After the substitution, the $u$ integral isn't something we want to deal with - so we switch the order and do the $z$ integral next. That's just a Gaussian - no elementary antiderivative, but we know that $int_0^infty e^-az^2,dz=fracsqrtpi2sqrta$. This leaves us with a $u$ integral that's algebraic; we substitute $v=sqrtu$, $dv=frac12sqrtu,du$ to rationalize it:
$$I = sqrtpiint_0^inftyfrac11+v^2,dv = sqrtpicdotfracpi2=fracpi^frac322$$
Done.

There's probably a way to connect this to a three-dimensional Gaussian, but I found it easier to break it down this way.