Find integers $1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$ Root numbers Problem (Math Quiz Facebook):

Consider the following equation:
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
Where $a,\,b,\,c,\,d$ are integers. Find $a+b+c+d$

I've tried it like this:
Let $w=\sqrt6,\, x=\sqrt3, \, y=\sqrt2, z=1$
$$\begin{align}
(y+z)^2 &= (y^2 + z^2) + 2yz\\
y+z &= \sqrt{(y^2 + z^2) + 2yz}\\
y+z &=  \sqrt{3 + \sqrt{8}}
\end{align}$$
Let $y+z=f$
$$\begin{align}
(x+f)^2 &= (x^2 + f^2) + 2xf\\
x+f &= \sqrt{(x^2 + f^2) + 2xf}\\
x+f &=  \sqrt{(9+\sqrt8) + 2\sqrt{9+3\sqrt8}}
\end{align}$$
And I don't think this going to work since there's still a root term on the bracket that is $9+\sqrt8$. I need another way to make it as an integer.
 A: First an answer $$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+ \sqrt{16588800}}}}.$$
Then an explanation.
Everything takes place inside the field $L=\Bbb{Q}(\sqrt2,\sqrt3)$. By elementary Galois theory the quadratic subfields of $L$ are $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt3)$ and $\Bbb{Q}(\sqrt6)$. The number $c+\sqrt d$ must be an element of $L$, so we can conclude that $d=\ell^2 e$ with some integer $\ell$ and $e\in \{2,3,6\}$ with the choice of $e$ depending on a circumstance we don't know yet.
The key question is the following:

Which elements of $L$ have squares in the subfield $\Bbb{Q}(\sqrt e)$?

The answer is, again by elementary Galois theory, that for example the square of a number $z=(a+b\sqrt2+c\sqrt3+d\sqrt6)$ is in $\Bbb{Q}(\sqrt6)$ if and only if either $a=d=0$ or $b=c=0$. Similarly with the other intermediate fields. This comes from the relevant automorphism of $L$ needing to have $z$ as an eigenvector belonging to one of the eigenvalues $+1$ or $-1$.
Let $\alpha=1+\sqrt2+\sqrt3+\sqrt6$. Then
$$
\alpha^2=12+8\sqrt2+6\sqrt3+4\sqrt6.
$$
In view of the previous observation we need to find integers $m,n$ such that $(\alpha^2-m)^2-n$ only contains terms with two of the alternative square roots.
Expanding gives
$$
(\alpha^2-m)^2=m^2-8 \sqrt{6} m-12 \sqrt{3} m-16 \sqrt{2} m-24 m+192 \sqrt{6}+272 \sqrt{3}+336 \sqrt{2}+476.$$
We need one of the square roots to disappear from this by careful choice of $m$.
Because $12\nmid 272$ we cannot make $\sqrt3$ disappear. The choice $m=24$ would make $\sqrt6$ disappear, but then we need to choose $n=476$ to kill the coefficient of $1$. The catch is that then $(\alpha^2-24)^2-476<0$ which is killjoy. It would lead to the answer
$$\alpha=\sqrt{24+\sqrt{476-\sqrt{5376+1536 \sqrt{6}}}},$$
but the negative square root is disallowed, I think.
Therefore we must kill the $\sqrt2$-terms from $(\alpha^2-m)^2$. This forces the choice $m=21$, when
$$
(\alpha^2-21)^2=413+20\sqrt3+24\sqrt6.
$$
This, in turn, forces $n=413$. As the last step we calculate
$$
(20\sqrt3+24\sqrt6)^2=4656+2880\sqrt2=4656+\sqrt{16588800}.
$$
A: Expand out enough to get to
\begin{align*}
(a^2-24a+476-b)+\sqrt{2}(336-16a)+\sqrt{3}(272-12a)+\sqrt{6}(192-8a)&=\sqrt{c+\sqrt{d}}.
\end{align*}
This means, when we square the left side, we need to only have two terms with nonzero coefficient. Note that
$$(w+x\sqrt2+y\sqrt3+z\sqrt6)^2=(w^2+2x^2+3y^2+6z^2)+2\sqrt2(wx+3yz)+2\sqrt3(wy+2xz)+2\sqrt6(wz+xy),$$
so we need two of $\{wx+3yz,wy+2xz,wz+xy\}$ to be $0$. However, if the first two are $0$, then
$$wxy+3y^2z=wxy+2x^2z=0$$
implies that either $z=0$ or $x=y=0$; in the first case, $w=0$. We may get similar conclusions for each of the other selections to be $0$, so we must have that two of the parameters $\{w,x,y,z\}$ are $0$. In particular, since none of our polynomials in $a$ for $x,y,z$ have common roots, we must have that $w=0$. Then, $y\neq 0$ since $y$ has a noninteger root for $a$, so we have $a\in\{21,24\}$ and $a=21\implies b=413$, with $a=24\implies b=476$. If $a=24$, the left side is actually negative (it's $-48\sqrt2-16\sqrt3$), so it can't be the square root of anything. For $a=21$, $b=413$, we may find by direct calculation that
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+\sqrt{16588800}}}}.$$
