is the number algebraic? Is the number $\alpha=1+\sqrt{2}+\sqrt{3}$ algebraic?
My first attempt was to try a polynomial for which $p(\alpha)=0$ for some $p(x)=a_{0}+a_{1}b_{1}+\cdots +b_{n-1}x^{n-1}$ i. e $x=1+\sqrt{2}+\sqrt{3}$ and then square it many times to get rid of the irrationals. This procedure was futile. 
Second attempt: I remember from the lecture we had that if $L=K(\alpha, \beta)$ with $\alpha,\beta$ algebraic over $K$ then $[L:K]<\infty$ moreover $[K(\gamma):K]<\infty$ for $\gamma=\alpha\pm \beta$ and $\gamma=\alpha\beta$ and $\gamma=\frac{\alpha}{\beta},\beta\neq 0$ as $K(\gamma)\subseteq L$ hence $\gamma$ is algebraic over $K$.
So applying the above to the problem then $\alpha=1+\sqrt{2}+\sqrt{3}$ is algebraic then over $K$ Right?  Or is there any other way to show it?
But I would like to know if it is possible to find a minimal polynomial having $\alpha$ as a zero? How to do that? 

$$\alpha=1+\sqrt2+\sqrt3$$ $$\alpha-1=\sqrt2+\sqrt3$$ $$(\alpha-1)^2=5+2\sqrt6$$ $$(\alpha-1)^2-5=2\sqrt6$$ $$((\alpha-1)^2-5)^2=24$$
