<s>[INST] triple integral $\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$ I need to solve this
triple integral $$\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$$ 
where $V$ is all $\Bbb R^3$ 
I've spent on this task a few hours,
ok firstly
$$e^{-x^2-y^2-z^2+xy+yz+xz}=e^{-1/2[(x-y)^2+(y-z)^2+(z-x)^2]}$$
I tried to use theorm of the change of variables, but
this substitution doesn't make sense
$$x-y=a$$
$$y-z=b$$
$$z-x=c$$
also
$$e^{-x^2-y^2-z^2+xy+yz+xz}=e^{(x-1/2y^2-1/2z^2)^2+((\sqrt{3}/2)y-
(\sqrt3/2)z)^2}$$
Maybe here change of variables'll be OK.
Please, help me. It's my homework, so important and I have no idea what I should do next.
 [/INST] To pursue the change-of-variables method further, note that with $a$ and $b$ defined as originally done, we have
$$
x^2 + y^2 + z^2 - xy - xz - yz = \frac{1}{2} \left[ (x - y)^2 + (y - z)^2 + (z - x)^2 \right] = \frac{1}{2} \left[ a^2 + b^2 + (a + b)^2 \right],
$$
since $a + b = x - z$.  Rearranging, we then get
$$
\frac{1}{2} \left[ a^2 + b^2 + (a + b)^2 \right] = \frac{1}{4} \left[ 3 (a + b)^2 + (a - b)^2 \right],
$$
since $a^2 + b^2 = \frac{1}{2}(a + b)^2 + \frac{1}{2}(a - b)^2$.  We can therefore use a different set of coordinates:
$$
u = a + b = x - z, \qquad v = a - b = x - 2y + z, \qquad w = z
$$
(Really, $w$ can be pretty much anything—see below.)  The integral then becomes, up to an overall constant coming from the Jacobian,
$$
\iiint e^{-\frac{3}{4}u^2 - \frac{1}{4} v^2} \, du \, dv\, dw.
$$
Since there is no $w$-dependence in the integrand, and since its range of integration will run from $-\infty$ to $\infty$, we conclude that the integral does not converge.
The dummy coordinate $w$, by the way, can be chosen to be any coordinate that linearly dependent on $x, y, z$ and that is not degenerate with $u$ and $v$ (i.e., one that doesn't cause the Jacobian matrix to have zero determinant, which basically means something that is not a linear combination of $u$ and $v$.) 
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