Let p(x) = a_0 + a_1 x + ldots + a_n x^n be a | vedclass

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(C) Let $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$.Squaring both sides:$x^2 = 2 + 3 + 6 + 2(\sqrt{6} + \sqrt{12} + \sqrt{18}) = 11 + 2(\sqrt{6} + 2\sqrt{3}

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$4$

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(C) Let $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$.Squaring both sides:$x^2 = 2 + 3 + 6 + 2(\sqrt{6} + \sqrt{12} + \sqrt{18}) = 11 + 2(\sqrt{6} + 2\sqrt{3} + 3\sqrt{2})$.Alternatively,consider $x - \sqrt{6} = \sqrt{2} + \sqrt{3}$.Squaring both sides:$(x - \sqrt{6})^2 = (\sqrt{2} + \sqrt{3})^2$$x^2 - 2\sqrt{6}x + 6 = 2 + 3 + 2\sqrt{6}$$x^2 + 1 = 2\sqrt{6}(x + 1)$.Squaring again:$(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$.Since $p(x)$ has integer coefficients and $p(\sqrt{2} + \sqrt{3} + \sqrt{6}) = 0$,the minimal polynomial of $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{6}$ over $\mathbb{Q}$ must divide any such $p(x)$.The degree of the minimal polynomial is $4$. Thus,the smallest possible value of $n$ is $4$.

Let $p(x) = a_0 + a_1 x + \ldots + a_n x^n$ be a non-zero polynomial with integer coefficients. If $p(\sqrt{2} + \sqrt{3} + \sqrt{6}) = 0$,then the smallest possible value of $n$ is

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