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?

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
$$\frac{4(1-2-3)}{1-4-6+1}=\frac{-16}{-8}=2.$$
