<s>[INST] Systems of Equations degree 2 Given the system 
$$\begin{aligned}
x^2+y^2+\frac{\sqrt 3}{2}xy&=32\\
x^2+z^2+\frac{1}{2}xz&=16\\
y^2+z^2&=16,\end{aligned}$$
find the value of $(xy+\sqrt{3}xz+2yz).$
The answer is $32,$ so I think there will be a really nice solution to this.
 I tried completing the square for each pair $x$ and $y$ but I could not find anything, I tried adding all three equations but I got nothing.
 [/INST] I guess the two factors (1/2) shouldn't be there, which mean we have instead
\begin{align}
x^2+y^2+\sqrt{3}xy&=32\tag{A}\\
x^2+z^2+xz&=16\tag{B}\\
y^2+z^2&=16\tag{C}.
\end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
\begin{align}
(x+\frac{\sqrt3}2y)^2+\frac14y^2&=32\\
(x+\frac12z)^2+\frac34z^2&=16\\
\end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=\left(8 \sqrt{\frac{7-2\sqrt{3}}{37}},
4(3 + \sqrt3)\sqrt{\frac{7-2\sqrt3}{111}},
4\sqrt{\frac{21 - 6\sqrt3}{37}}\right),\text{ and similar}
$$
and it is easy to check $(xy+\sqrt{3}xz+2yz)^2=2^{10}$.  Indeed, it follows immediately from noting the LHS of
$$
\mathrm{(A)^2+(B)^2+(C)^2-2\times(A)\times(B)-2\times(B)\times(C)-2\times(C)\times(A)}
$$
is $-(xy+\sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.
</s>