I am interested in the following question:
If $\sqrt{2}+\sqrt{3}+\sqrt{6}$ is a root of $x^4+ax^3+bx^2+cx+d=0$, where $a,b,c,d\in\mathbb{Z}$, then find the value of $|a+b+c+d|$.
I tried substituting this root in the polynomial to take the irrational part one side and the rational part the other side, but that became very messy. I am in search of an elegant method. Any help will be highly appreciated.
On
Let $y=\sqrt 2+\sqrt 3+\sqrt 6$
You can compute directly as follows:
Note first that $y^2=\left(\sqrt 2+\sqrt 3+\sqrt 6\right)^2=2+3+6+2\sqrt 6+6\sqrt 2+4\sqrt 3=11+2y+4\sqrt 2+2\sqrt 3$
Then $y^3=11y+2y^2+(4\sqrt 2+2\sqrt 3)y$ and we work with the last term
$y^3=11y+2y^2+14+6\sqrt 6+8\sqrt 3+6\sqrt 2=14+17y+2y^2+2\sqrt3$ and
$y^4=14y+17y^2+2y^3+2\sqrt 3 y=14y+17y^2+2y^3+2\sqrt 6+6+6\sqrt 2$
Then we have $2\sqrt 6+6\sqrt 2=y^2-11-4\sqrt 3$
and $2\sqrt 3= y^3-2y^2-17y-14$ which can be used to eliminate the radicals, leaving the explicit quartic you want.
I reckon the four roots of the quartic will be $\sqrt2+\sqrt3+\sqrt6$, $\sqrt2-\sqrt3-\sqrt6$, $-\sqrt2+\sqrt3-\sqrt6$ and $-\sqrt2-\sqrt3+\sqrt6$ since these are the Galois conjugates of $\sqrt2+\sqrt3+\sqrt6$. So, $$1+a+b+c+d=(1-\sqrt2-\sqrt3-\sqrt6)(1-\sqrt2+\sqrt3+\sqrt6)(1+\sqrt2-\sqrt3+\sqrt6)(1+\sqrt2+\sqrt3-\sqrt6)$$ etc.