Minimum degree rational equation with root $a+\sqrt{b}+\sqrt{c}+\sqrt{d}$. What is the minimum degree of an equation with rational coefficients that has a root $x=a+\sqrt{b}+\sqrt{c}+\sqrt{d}$ with $a,b,c,d$ primes numbers?
I know how to find an equation of second degree that has root $a+\sqrt{b}$
$$
x=a+\sqrt{b} \quad \rightarrow \quad (x-a)^2=b 
$$
and a $4-$degree equation that has root $a+\sqrt{b}+\sqrt{c}$ 
$$
x=a+\sqrt{b}+\sqrt{c} \quad \rightarrow \quad (x-a)^2=(\sqrt{b}+\sqrt{c})^2 \quad \rightarrow \quad  \left[(x-a)^2-b-c \right]^2=4bc
$$
But it seems that this simple  method cannot be used for a root with more than two surds.  There is it some other method?

Let $x=a+ \sqrt{b}+ \sqrt{c}+ \sqrt{d}$  square this and move some terms around
\begin{eqnarray*}
(x-a)^2+b-c-d = -2(x-a)\sqrt{b}+ 2\sqrt{c} \sqrt{d}
\end{eqnarray*}
Square it again and move some more terms around
\begin{eqnarray*}
((x-a)^2+b-c-d)^2 -4b(x-a)^2-4cd = -8(x-a)\sqrt{b}\sqrt{c} \sqrt{d}
\end{eqnarray*}
Squaring one final time & we have 
\begin{eqnarray*}
(((x-a)^2+b-c-d)^2 -4b(x-a)^2-4cd)^2 = 64(x-a)^2bcd
\end{eqnarray*}
So the equation that this quantity satisfies an equation of degree $\color{red}{8}$ as expected.
