Rationalise Denominator: 1/(1 + √2 + √3) by using: (1 - √2 + √3) as conjugate?

`k = 1/(1 + {√2 + √3})`

`rArr k = 1/({√3+1}+√2)`

`rArrk = ({√3+1}-√2)/[({√3+1}+√2).({√3+1}-√2)]`

`rArrk = ({√3+1}-√2)/[(√3+1)^2-(√2)^2]`

`rArrk = ({√3+1}-√2)/[3+1+2√3-2]`

`:.k = ({√3+1}-√2)/[2(1+√3)]`

The difference of squares identity can be written:

`A^2-B^2=(A-B)(A+B)`

This is the key to eliminating square roots from the denominator.

Note that `(1-sqrt(2)+sqrt(3))` is only a partial conjugate for `(1+sqrt(2)+sqrt(3))`. Multiplying these two expressions will eliminate terms in `sqrt(2)` but leave terms in `sqrt(3)`.

If we want to rationalise the denominator, we will also need to multiply by some expression of the form `a+bsqrt(3)` (actually I will use `(sqrt(3)-1)`)

`1/(1+sqrt(2)+sqrt(3))`

`=((1+sqrt(3))-sqrt(2))/(((1+sqrt(3))-sqrt(2))((1+sqrt(3))+sqrt(2)))`

`=(1+sqrt(3)-sqrt(2))/((1+sqrt(3))^2-(sqrt(2))^2)`

`=(1+sqrt(3)-sqrt(2))/(1+2sqrt(3)+3-2)`

`=(1+sqrt(3)-sqrt(2))/(2(sqrt(3)+1))`

`=((sqrt(3)-1)(1+sqrt(3)-sqrt(2)))/(2(sqrt(3)-1)(sqrt(3)+1))`

`=((sqrt(3)+3-sqrt(6))-(1+sqrt(3)-sqrt(2)))/(2(3-1))`

`=(2+sqrt(2)-sqrt(6))/4`

Note that having got to this result, the numerator is a proper conjugate for `1+sqrt(2)+sqrt(3)` in that:

`(1+sqrt(2)+sqrt(3))(2+sqrt(2)-sqrt(6)) = 4`