Simplification of $\frac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$ 
Simplify
$$\dfrac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$$
Final solution should have rational denominators.

Suppose the solution is $X$, I have tried to make up an equation for $X^2$
$$X^2 = \frac{200}{-2\sqrt{15}+2\sqrt{35}-\sqrt{21}+10}$$
My idea is that solving for $X^2$, if can be simply done, can easily give us $X$.
Please help!
 A: Hint: we are generalising the method of rationalising a denominator of two terms. For instance, if our denominator was $\sqrt{a}-b$ then we would multiply (numerator and denominator) by $\sqrt{a}+b$ to get $a-b^2$. That is, we are using "difference of two squares".
Observe that $[(\sqrt{7}+2\sqrt{5})-\sqrt{3}][(\sqrt{7}+2\sqrt{5})+\sqrt{3}]=(\sqrt{7}+2\sqrt{5})^2-(\sqrt{3})^2=7+4\sqrt{35}+20-3$
Now what can we do?
A: For a shortcut, let $\,a=\sqrt{3}, b=\sqrt{5}, c=\sqrt{7}\,$, then:
$$
\require{cancel}
\begin{align}
& \dfrac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt{7}+2\sqrt{5}-\sqrt{3}} = \frac{2ac-bc+5ab-16}{c+2b-a}
\\ &\quad\quad\quad\quad =\frac{c(2a-b)\color{red}{+(2b-a)(2a-b)-(2b-a)(2a-b)}+5ab-16}{c+2b-a}
\\ &\quad\quad\quad\quad = \frac{(c+2b-a)(2a-b)-\cancel{5ab}+2a^2+2b^2+\cancel{5ab}-16}{c+2b-a}
\\ &\quad\quad\quad\quad = 2a-b+\frac{\cancel{6 + 10 - 16}}{c+2b-a}
\end{align}
$$
A: The solution below is a lot of work to do by hand but at least it does not need any bright ideas for algebraic manipulation.
The problem reduces to finding $\dfrac{1}{\sqrt7+2\sqrt5-\sqrt3}$.
Now, $\theta=\sqrt7+2\sqrt5-\sqrt3$ is a root of $x^8 - 120 x^6 + 3632 x^4 - 28800 x^2 + 256$. The extended Euclidean algorithm gives
$$
1 = \frac{1}{256}(x^8 - 120 x^6 + 3632 x^4 - 28800 x^2 + 256)+ \frac{1}{256}(-x^7 + 120 x^5 - 3632 x^3 + 28800 x)x
$$
Therefore,
$$
\dfrac{1}{\sqrt7+2\sqrt5-\sqrt3} = \frac{1}{\theta} =\frac{1}{256}(-\theta^7 + 120 \theta^5 - 3632 \theta^3 + 28800 \theta)
$$
A: The solution below can be done by hand without any bright ideas.
We seek a solution of the form
$$
\theta=q_1 + q_2 \sqrt{3} + q_3 \sqrt{5} + q_4 \sqrt{7} +q_5 \sqrt{15} +q_6 \sqrt{21} +q_7 \sqrt{35} + q_8 \sqrt{105}
$$
with $q_1,q_2,\dots,q_8 \in \mathbb Q$.
Then
$$
(2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16)=\theta(\sqrt7+2\sqrt5-\sqrt3)
$$
becomes a sparse linear system in $q_1,q_2,\dots,q_8$, whose solution is
$$
q_1 = 0, q_2 = 2, q_3 = -1, q_4 = 0, q_5 = 0, q_6 = 0, q_7 = 0, q_8 = 0
$$
The main point here is that this problem lives in the field $K=\mathbb Q(\sqrt{3},\sqrt{5},\sqrt{7})$ and that $1,\sqrt{3},\sqrt{5},\sqrt{7}, \sqrt{15},\sqrt{21},\sqrt{35},\sqrt{105}$ is a basis for $K$ over $\mathbb Q$.
If you don't know this bit of field theory, you can still use this solution because the $\theta$ above works.
