evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$ 
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$

How to evalute this equation without using calculator?
 A: Another way : $$\begin{align}\\&\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\\&=\left(\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}\right)+\left(\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\right)\\&=\frac{1+\sqrt 2-\sqrt 3+1+\sqrt 2+\sqrt 3}{(1+\sqrt 2+\sqrt 3)(1+\sqrt 2-\sqrt 3)}+\frac{1-\sqrt 2-\sqrt 3+1-\sqrt 2+\sqrt 3}{(1-\sqrt 2+\sqrt 3)(1-\sqrt 2-\sqrt 3)}\\&=\frac{2+2\sqrt 2}{(1+\sqrt 2)^2-3}+\frac{2-2\sqrt 2}{(1-\sqrt 2)^2-3}\\&=\frac{2+2\sqrt 2}{2\sqrt 2}+\frac{2-2\sqrt 2}{-2\sqrt 2}\\&=2\end{align}$$
A: Add the first two
$$\frac 1{1+\sqrt3+\sqrt2} + \frac 1{1+\sqrt3-\sqrt2}=2\frac{1+\sqrt3}{(1+\sqrt3)^2-2}=2\frac{1+\sqrt3}{2+2\sqrt3}=1.$$
Similarly, add the last two 
$$2\frac{1-\sqrt3}{2-2\sqrt3}=1.$$
A: We have, $$\underbrace{\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}} + \underbrace{\frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}}$$ $$=\left(\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}\right) +\left( \frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}\right)$$ $$=\left(\frac {1-\sqrt2+\sqrt3+1+\sqrt2+\sqrt3}{(1+\sqrt2+\sqrt3)(1-\sqrt2+\sqrt3)} \right) +\left( \frac {1-\sqrt2-\sqrt3+1+\sqrt2-\sqrt3}{(1+\sqrt2-\sqrt3)(1-\sqrt2-\sqrt3)} \right)$$ $$=\left(\frac {2+2\sqrt3}{(1+\sqrt3)^2-(\sqrt{2})^2} \right) +\left( \frac {2-2\sqrt{3}}{(1-\sqrt3)^2-(\sqrt{2})^2} \right)$$   $$=\frac{2+2\sqrt{3}}{1+3+2\sqrt{3}-2}+\frac{2-2\sqrt{3}}{1+3-2\sqrt{3}-2}$$ $$=\frac{2+2\sqrt{3}}{2+2\sqrt{3}}+\frac{2-2\sqrt{3}}{2-2\sqrt{3}}$$ $$=1+1=2$$
A: We have that $\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial $(x^2-5)^2-24$, hence:
$$ x^4-10\,x^2+1 = \prod_{\xi_i\in Z}(x-\xi) $$
and $1\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial:
$$ (x-1)^4-10(x-1)^2+1 = x^4-4x^3-4x^2+16x-8. $$
By Viète's theorem, the sum of the roots of a polynomial $p(x)$ raised to the minus one power is given by $-\frac{[x^1]\,p(x)}{[x^0]\,p(x)}$, hence the answer is $-\frac{16}{-8} = \color{red}{2}$.
A: from here,  separating it so that they have same thing
$$\frac 1{1+(\sqrt2+\sqrt3)} + \frac 1{1-(\sqrt2-\sqrt3)} + \frac 1{1+(\sqrt2-\sqrt3)} + \frac 1{1-(\sqrt2+\sqrt3)}$$
let $x = \sqrt 2 + \sqrt 3$, $y = \sqrt 2 - \sqrt 3$
$$\frac 1{1+x} + \frac 1{1-y} + \frac 1{1+y} + \frac 1{1-x}$$
combine fraction with x into one fraction, same thing as y
$$=\frac {(1+x)+(1-x)}{(1+x)(1-x)} + \frac {(1+y)+(1-y)}{(1+y)(1-y)}$$
$$=\frac {2}{1-x^2} + \frac {2}{1-y^2}$$
let them become one fraction
$$=\frac {2(1-x^2)+2(1-y^2)}{(1-x^2)(1-y^2)}$$
$$=\frac {4-2x^2-2y^2}{1-x^2-y^2+x^2y^2}$$
calculate $x^2$ and $y^2$, $x^2=(\sqrt 2 + \sqrt 3)^2= 2 + 2(\sqrt 6) + 3 = 2 + \sqrt {6\times4} + 3 = 5 + \sqrt {24}$, $y^2=(\sqrt 2 - \sqrt 3)^2= 2 - 2(\sqrt 6) + 3 = 2 - \sqrt {6\times4} + 3 = 5 - \sqrt {24}$
now replace $x^2$ and $y^2$
$$=\frac {4-2(5 + \sqrt {24}) - 2(5-\sqrt {24})}{1-(5+\sqrt {24})-(5-\sqrt {24})+((5+\sqrt {24})(5-\sqrt {24}))}$$
$$=\frac {4-10 - 2\sqrt {24} - 10+2\sqrt {24}}{1-5-\sqrt {24}-5+\sqrt {24}+(25-24)}$$
$$=\frac {-16}{-8} = 2$$
A: Rearranging, we have
$$\frac{1}{1+(a+b)}+\frac{1}{1-(a+b)}+\frac{1}{1-(a-b)}+\frac{1}{1+(a-b)},$$
giving
$$\frac{2}{1-(a+b)^2}+\frac{2}{1-(a-b)^2}.$$
Then,
$$\frac{4-2(a-b)^2-2(a+b)^2}{1-(a-b)^2-(a+b)^2+(a+b)^2(a-b)^2}.$$
Expanding,
$$\frac{4(1-a^2-b^2)}{1-2a^2-2b^2+(a^2-b^2)^2}.$$
In your case
$a=\sqrt{2}$ and $b=\sqrt{3}$, so that your sum is
$$\frac{4(1-2-3)}{1-4-6+1}=\frac{-16}{-8}=2.$$
