The irrational biquadratic root

First note that $x+1=\left(\sqrt2+1\right)\left(\sqrt3+1\right)$ , so $\begin{aligned} \left(\frac{x+1}{\sqrt2+1}-1\right)^2&=3 \\ \left((x+1)-\left(\sqrt2+1\right)\right)^2&=3\left(\sqrt2+1 \right)^2 \\ \left(x-\sqrt2\right)^2&=3\left(\sqrt2+1 \right)^2 \\ x^2-2\sqrt2 x+2&=9+6\sqrt2 \\ x^2-7 &= 6\sqrt2+2\sqrt2 x \\ x^4-14x^2+49 &= 72+48x+8x^2 \\ x^4-22x^2-48x-23&=0\end{aligned}$

so that $|a+b+c+d|=\boxed{93}$ .

Note that we need the uniqueness of minimal polynomials to be sure we have the right answer here.

Also note that the roots of this polynomial are $\sqrt2+\sqrt3+\sqrt6,\;\;\sqrt2-\sqrt3-\sqrt6,\;\;-\sqrt2+\sqrt3-\sqrt6,\;\;-\sqrt2-\sqrt3+\sqrt6$

so I'm sure there's a neater (Galois theory) approach to this, though less sure how easy it is to explain.

$\begin{aligned} x & = \sqrt 2 + \sqrt 3 + \sqrt 6 \\ x(x - 2\sqrt 6) & = (\sqrt 2 + \sqrt 3 + \sqrt 6)(\sqrt 2 + \sqrt 3 - \sqrt 6) \\ x^2 - 2\sqrt 6x & = 2\sqrt 6 - 1 \\ x^2 + 1 & = 2\sqrt 6(x+1) \\ (x^2+1)^2 & = 24(x+1)^2 \\ x^4 + 2x^2 + 1 & = 24(x^2 + 2x + 1) \\ x^4 - 22x^2 - 48x - 23 & = 0 \end{aligned}$

Therefore $|a+b+c+d| = |0-22-48-23| = \boxed{93}$ .

Let $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{6}.$ Since $\alpha \in {\mathbb Q}(\sqrt{2},\sqrt{3}),$ the conjugates of $\alpha$ are its images under the elements of the Galois group of this field. These send $\sqrt{2}$ to $\pm \sqrt{2}$ and $\sqrt{3}$ to $\pm \sqrt{3}$ (which determines where $\sqrt{6}$ goes). The four conjugates are $\sqrt{2} + \sqrt{3} + \sqrt{6}, \sqrt{2} - \sqrt{3} - \sqrt{6}, -\sqrt{2} + \sqrt{3} - \sqrt{6}, -\sqrt{2} - \sqrt{3} + \sqrt{6}.$

So the minimal polynomial of $\alpha$ is equal to the product of $(x-r_i),$ where the $r_i$ are the four conjugates of $x.$ Grouping appopriately, we get $\begin{aligned} ((x-\sqrt{6})^2-(\sqrt{2}+\sqrt{3})^2)((x+\sqrt{6})^2-(\sqrt{2}-\sqrt{3})^2) &= (x+\sqrt{6})^2(x-\sqrt{6})^2 - (x+\sqrt{6})^2(5+2\sqrt{6}) - (x-\sqrt{6})^2(5-2\sqrt{6}) + 1 \\ &= (x^2-6)^2 + 1 - (10x^2 + 48x + 60) \\ &= x^4 - 22x^2 - 48x -23 \end{aligned}$ so the answer is $\fbox{93}.$