How can you write the expression with rationalized denominator? sqrt3 - sqrt6 / sqrt 3 + sqrt 6

sqrt 3 - sqrt 6 / sqrt 3 + sqrt 6
can be written as 

sqrt 3 + sqrt 6 - (sqrt 6 / sqrt 3)

I just did some rearranging, no changes yet

Anything times 1 is itself, but the fraction 8/8 is also = 1, and sqrt 67 / sqrt 67 is also 1, so we can multiply anything by anything as long as the number is divided by itself

so we can say

sqrt 3 + sqrt 6 - (sqrt 6 / sqrt 3) = 
sqrt 3 + sqrt 6 - (sqrt 6 / sqrt 3) * (sqrt 3 / sqrt 3)

Now if we write it vertically so its easier to see it becomes

                              (sqrt 6)       (sqrt 3)
 sqrt 3 + sqrt 6  -       *    = the original equation
                              (sqrt 3)       (sqrt 3)

on the denominator, sqrt 3 * sqrt 3 AKA (sqrt 3) squared = 3 because the square and the square root cancel out

but the numerator is (sqrt 6) * (sqrt 3) and that's equal to sqrt 18

so we can take a perfect square out of that
18 = 9 * 2 and 9 is a perfect square because
3 * 3 = 9

so we can say that
sqrt 18 = sqrt 9 * sqrt 2 
and sqrt 9 = 3 so
sqrt 18 = 3 * sqrt 2

which is cool because our denominator was sqrt 3 * sqrt 3 which is equal to 3 which now cancels out! So now we have 

                             (sqrt 6)       (sqrt 3)
 sqrt 3 + sqrt 6  -       *    = 
                              (sqrt 3)       (sqrt 3)

                             3 (sqrt 2) 
 sqrt 3 + sqrt 6  -      
                                 3     
3's cancel out and we get

sqrt 3 + sqrt 6 - sqrt 2

The variables should add up to equal 1