Solve in real numbers the system of equations $\{\begin{array}{l}(x-2y)(x-4z)=3\\ (y-2z)(y-4x)=5\\ (z-2x)(z-4y)=-8.\end{array}$ Then find the value o$x^2+y^2+z^2$?

Since there is no apparent symmetry in this system, we start by brutally multiplying out the factors in the left hand-side and adding the resulting equations (one hint being furnished by the fact that $3+5-8=0$and that all terms in the left hand-side are homogeneous).
We end up with
$x^2+y^2+z^2+2zx+2zy+2yx=0,$
which is equivalent to  $(x+y+z{)}^2=0$ and then to x=-y-z. Replacing this value of x in the first and the second equations, we obtain the system in the unknowns y,z
$\{\begin{array}{l}3y^2+16yz+5z^2=3\\ 5y^2-6zy-8z^2=5\end{array}$
Next, using that $3\cdot 5-5\cdot 3=0$, we multiply the first equation by 5 , the second one by 3 and we subtract the resulting relations. This gives
$29z^2+58zy=0,\text{that is }z(z+2y)=0\text{.}$

Let us discuss two cases. If z=0, then the previous system becomes $3y^2=3\text{and}5y^2=5$ having two solutions $\text{s}y=\pm 1\text{. Since}x=-y-z$ we obtain two solutions $(x,y,z)=\left\{\right(1,-1,0),(-1,1,0\left)\right\}$ of the initial system. If on the other hand z=-2y, then the first equation of the previous system becomes
$3y^2-32y^2+20y^2=3\text{,}$
that is $-9v^2=3\text{.}$ and this clearly has no real solution. We conclude that this second case is impossible, hence the system has only the solutions found above.