<s>[INST] Difficult denomiator rationalization questions These are two questions from a competitive exam involving irrationals where I am supposed to simplify it to match one of the given options.

QUESTION 1: The value of $$ \frac {2 (\sqrt 2+ \sqrt6)}{3(\sqrt {2 + \sqrt 3})} + \sqrt {2 + \sqrt 3}+ \sqrt {2 - \sqrt 3}$$ 
  is 
A. $\frac {3+4 \sqrt 6}{3}$
B. $\frac {4+3 \sqrt 6}{3}$
C. $\frac {3+4 \sqrt 6}{4}$
D. $\frac {4- 3\sqrt 6}{3}$

I have been able to solve this far:
Taking denominator common, I get
$$ \frac {2 (\sqrt 2+ \sqrt6) + 3(\sqrt {2 + \sqrt 3})(\sqrt {2 + \sqrt 3})+ 3(\sqrt {2 + \sqrt 3})\sqrt {2 - \sqrt 3}}{3(\sqrt {2 + \sqrt 3})}$$
After which:
$$ \frac {2 (\sqrt 2+ \sqrt6) + 3( {2 + \sqrt 3})+ 3(\sqrt {2^2 - \sqrt 3^2})}{3(\sqrt {2 + \sqrt 3})}$$
which gives
$$ \frac {2 (\sqrt 2+ \sqrt6) + 3( {2 + \sqrt 3})+ 3}{3(\sqrt {2 + \sqrt 3})}$$
I multiply by the conjugate of the irrational term  in denominator.
$$ \frac {2 (\sqrt 2+ \sqrt6)(\sqrt {2 - \sqrt 3}) + 3( {2 + \sqrt 3})\sqrt {2 - \sqrt 3}+ 3(\sqrt {2 - \sqrt 3})}{3(\sqrt {2 + \sqrt 3})(\sqrt {2 - \sqrt 3})}$$
Upon simplification,
$$ \frac {2 (\sqrt 2+ \sqrt6)(\sqrt {2 - \sqrt 3}) + 3(\sqrt 3)\sqrt {2 - \sqrt 3}+ 9(\sqrt {2 - \sqrt 3})}{3}$$
Beyond this, I am not able to work out a solution. Any hints (please explain the hint slightly) are welcome.

QUESTION 2: The value of
  $$ \sqrt {43-12 \sqrt 7} - \frac {2}{\sqrt {16+6 \sqrt 7}}$$
is:
A. $-3$
B. $3 $
C. $2 \sqrt 7 -3 $
D. $- (2 \sqrt 7 +3) $

I have been able to solve this far:
Taking denominator common:
$$ \frac {\sqrt {43-12 \sqrt 7}(\sqrt {16+6 \sqrt 7}) - 2}{\sqrt {16+6 \sqrt 7}}$$
Upon simplification I get:
$$ \frac {\sqrt {184 + 66\sqrt 7} - 2}{\sqrt {16+6 \sqrt 7}}$$
Multiplying by the conjugate of the denominator:
$$ \frac {\sqrt {184 + 66\sqrt 7}(\sqrt {16-6 \sqrt 7}) - 2(\sqrt {16-6 \sqrt 7)}}{\sqrt {16+6 \sqrt 7}(\sqrt {16-6 \sqrt 7)}}$$
Simplification gives:
$$ \frac {\sqrt {172 - 48\sqrt 7}- 2(\sqrt {16-6 \sqrt 7)}}{\sqrt 4}$$
When I typed this into a software I got 
$\sqrt {16-6 \sqrt 7} = 3 - \sqrt 7 $ and
$ \sqrt {172 - 48\sqrt 7} = 12 - 2 \sqrt7  $
How do we get these?
 [/INST] As per the request of the OP I am posting my comment as an answer.
If you square the expression $$\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}$$ you will find that this is $\sqrt{6}$. This method works for any expression of the form $$\sqrt{n+\sqrt{n^2-1}}+\sqrt{n-\sqrt{n^2-1}}.$$

In order to obtain a simpler version of rational expressions of the form $x=\sqrt{a+b\sqrt{7}}$ we write $$x^2=a+b\sqrt{7}$$ then $$(x^2-a)^2 = 7b^2$$ and so $x$ satisfies the expression $$0=(x^2-a)^2 - 7b^2=x^4-2ax^2 + (a^2 - 7b^2)$$
Next find a factorization for this polynomial, looking for the so called "minimal polynomial" for $x$. That will give you the simplification you desire.
The minimal polynomial corresponding to $\sqrt{43-12\sqrt{7}}$ is $x^2-12x+29$. This means $x = 6 \pm \sqrt{7}$. We conclude that $x=6-\sqrt{7}$ since $6+\sqrt{7} >7$ whose square is larger than 43.

You can factor this polynomial as follows: $$(x^2-a)^2 = 7b^2$$ gives $$x^2 - a = \pm \sqrt{7}b$$ which in turn gives $$x = \pm \sqrt{a \pm \sqrt{7}b}.$$ Thus the polynomial factors as $$\left(x-\sqrt{a - \sqrt{7}b}\right)\left(x-\sqrt{a + \sqrt{7}b}\right)\left(x+\sqrt{a - \sqrt{7}b}\right)\left(x+\sqrt{a + \sqrt{7}b}\right)$$ Picking clever combinations of these will yield the minimal polynomial.
For instance $$\left(x-\sqrt{a - \sqrt{7}b}\right)\left(x+\sqrt{a + \sqrt{7}b}\right)=x^2+x\left(\sqrt{a + \sqrt{7}b}-\sqrt{a - \sqrt{7}b}\right) + \sqrt{a^2-7b^2}.$$ Further by squaring $\sqrt{a + \sqrt{7}b}-\sqrt{a - \sqrt{7}b}$ we see that this is equal to $\sqrt{2a - 2\sqrt{a^2-7b^2}}$. If you are lucky, these will be integers. If not, try combining them differently multiplying terms with $\sqrt{a-\sqrt{7}b}$ against $\sqrt{a+\sqrt{7}b}$. If no combination yields quadratic terms with integer coeficients, there is no quadratic minimal polynomial and the expression cannot be simplified.
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