<s>[INST] How do I solve this system of equations (with squares and square roots)? Consider the system of equations:
\begin{align*}
x+y+z&=6\\
x^2+y^2+z^2&=18
\\\sqrt{x}+\sqrt{y}+\sqrt{z}&=4.
\end{align*}
How do I solve this? I've tried squaring, adding equations side by side, substituting, etc., but without success, e.g.
$$x^2+y^2+z^2+2(xy+yz+xz)=36\implies xy+yz+xz=9,$$
but then I don't know what to do next. Please help me solve this.
 [/INST] Let $s = \sqrt{x}$, $t = \sqrt{y}$, $u = \sqrt{z}$.  Then your system is
$$ \eqalign{s^2 + t^2 + u^2 &= 6\cr
            s^4 + t^4 + u^4 &= 18\cr
            s + t + u &= 4\cr}$$
Solving the third equation for $u$ and substituting in the others gives
$$ \eqalign{s^2+t^2+(4-s-t)^2-6 &= 0\cr s^4+t^4+(4-s-t)^4-18 &= 0\cr}$$
The resultant of the left sides with respect to $t$ is 
$$ 4096 (s-2)^2 (s-1)^4$$
So the only solutions have $s = 1$ or $s = 2$ (and by symmetry, also $t=1$ or $t=2$ and $u=1$ or $u=2$).  In fact the solutions all have two of the variables $=1$ and one $=2$.  In terms of $x,y,z$, two are $1$ and one is $4$.
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