If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$? From the Pre-Regional Mathematics Olympiad, 2019:

If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$?

I have provided one answer below, and would be interested in alternative solutions.

Using a bit of Galois theory: first, note that the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$, none of which contains $\sqrt{2} + \sqrt{3} + \sqrt{6}$.  Therefore, $\sqrt{2} + \sqrt{3} + \sqrt{6}$ is a primitive generator of the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ of $\mathbb{Q}$, so the degree of its minimal polynomial is $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4$.
Now, also note that if the minimal polynomial is $p(x) = x^4 + ax^3 + bx^2 + cx + d$, then $a + b + c + d = p(1) - 1$.  On the other hand, we also know that $p(x) = \prod_{\sigma \in G} (x - \sigma(\alpha))$ where $G = \operatorname{Gal}(\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q})$ and $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{6}$.  Therefore, $p(1) = N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}}(1 - \alpha)$.
But also,
$$
N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}} (1 - \alpha) = N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}(\sqrt{2})} [(1 - \sqrt{2}) - (1 + \sqrt{2}) \sqrt{3}] = \\
N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} [(1 - \sqrt{2})^2 - 3 (1 + \sqrt{2})^2] = N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} (-6 - 8 \sqrt{2}) = 6^2 - 2 \cdot 8^2 = -92.
$$
Thus, $a + b + c + d = -93$, and $|a + b + c + d| = 93$.
