Let $a_0+a_1x+....+a_nx^n$ be a non zero polynomial with integer coefficients.if $p(√2+√3+√6)=0$, the smallest possible value of n is?

Given the question is about a multiple choice test with possible answers $2,4,6,8$ maybe the most direct and elementary proof is to notice that, with $x=\sqrt 2 + \sqrt 3 + \sqrt 6\;$:

• $(x-\sqrt 6)^2=(\sqrt 2 + \sqrt 3)^2 \iff x^2+6-2x \sqrt 6 = 2 + 3 + 2 \sqrt 6\;$ then isolating $\sqrt 6$ on one side and squaring again we get a $4^{th}$ degree equation in $x$, which eliminates answers $n=6,8$.

• To eliminate $n=2$, note that $x^2=11 + 6 \sqrt 2 + 4 \sqrt 3 + 2 \sqrt6$. If an equation existed with rational coefficients $a x^2 + b x + c = 0$ then that would give a linear dependency between $\sqrt 2, \sqrt 3,\sqrt6$, but they are linearly independent over the rationals (see for example How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$?).

Which leaves the answer $n=4$.

$\alpha=\sqrt 2+\sqrt 3+\sqrt 6 \in \mathbb Q[\sqrt 2,\sqrt 3]$, which has degree $4$ over $\mathbb Q$. A basis is $1,\sqrt 2,\sqrt 3,\sqrt 6$.

So, the degree of $\alpha$ is at most $4$ and must be divisor of $4$.

The degree is not $1$, because $\alpha$ is not rational, since otherwise $1,\sqrt 2,\sqrt 3,\sqrt 6$ would be linearly dependent.

The degree is not $2$, because $\alpha \notin \mathbb Q[\sqrt 2]$ and $\alpha \notin \mathbb Q[\sqrt 3]$, since otherwise $1,\sqrt 2,\sqrt 3,\sqrt 6$ would be linearly dependent.

So, the only possibility left is that the degree is $4$.