Rationalize the denominator:
√2 + √3 + √6
-------------------------
√2 + √3 + √6 + √8 + √16
The denominator can be factored, and the numerator is almost factorizable.
sqrt(2)+sqrt(3)+sqrt(6)
= [1+sqrt(2)+sqrt(3)+sqrt(2)*sqrt(3)] - 1
= [1+sqrt(2)]*[1+sqrt(3)] - 1
sqrt(2)+sqrt(3)+sqrt(6)+sqrt(8)+sqrt(16)
= 3*sqrt(2)+4+sqrt(3)+sqrt(6)
= 3*sqrt(2)+3*sqrt(2)*sqrt(2)+sqrt(3)+sqrt(3)*sqrt(2)
= [3*sqrt(2)+sqrt(3)]*[1+sqrt(2)]
= [sqrt(3)]*[1+sqrt(6)]*[1+sqrt(2)]
Then the whole fraction can be written as
[1+sqrt(2)]*[1+sqrt(3)] - 1
---------------------------------------
[sqrt(3)]*[1+sqrt(6)]*[1+sqrt(2)]
To keep things clear I will rationalize the denominator one factor at a time. First multiply the numerator and denominator by sqrt(2)-1
sqrt(3)-sqrt(2)
------------------------
[sqrt(3)]*[1+sqrt(6)]
Then multiply the numerator and denominator by sqrt(6)-1
3*sqrt(2)-2*sqrt(3)-sqrt(3)+sqrt(2)
----------------------------------------
5*sqrt(3)
4*sqrt(2)-3*sqrt(3)
= ----------------------
5*sqrt(3)
Then finish with cleaning up sqrt(3) for a final answer of
(4*sqrt(6) - 9) / 15.
Solution
