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?

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 $y^2$
$$=\frac {4-2(5 + \sqrt {24}) - 2(5-\sqrt {24})}{1-(5+\sqrt {24})-(5-\sqrt {24})+((5+\sqrt {24})(5-\sqrt {24}))}$$
$$=\frac {4-10 - 2\sqrt {24} - 10+2\sqrt {24}}{1-5-\sqrt {24}-5+\sqrt {24}+(25-24)}$$
$$=\frac {-16}{-8} = 2$$
