Question

How do you rationalize the denominator and simplify `1/{1+sqrt(3)-sqrt(5)}`?

Answer 1

Attempt to make the three term denominator two terms..

Explanation:

Multiply the fraction by `(1+ sqrt3 + sqrt5)/(1+sqrt3+sqrt5)` . This is the same as multiplying by 1. This is two eliminate two surds.
The denominator you should get after multiplying the two fractions is
`(1+sqrt3+sqrt5) * (1+sqrt3-sqrt5)` . We can write this as `((1+sqrt3)-sqrt5)(1+sqrt3)+sqrt5)`.
This can then be made is DOPS and written as : `(1+sqrt3)^2-(sqrt5)^2`.
Simplify that to get: `1+3+2sqrt3-5`
I'll leave it here :).

Answer 2

`1/(1+sqrt(3)-sqrt(5)) =(7+3sqrt(3)+sqrt(5)+2sqrt(15))/11`

Explanation:

This involves two stages of rationalisation to get rid of terms in `sqrt(3)` and `sqrt(5)`. Both steps use the difference of squares identity:

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

So:

`1/(1+sqrt(3)-sqrt(5)) = (1+sqrt(3)+sqrt(5))/(((1+sqrt(3))-sqrt(5))((1+sqrt(3))+sqrt(5)))`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = (1+sqrt(3)+sqrt(5))/((1+sqrt(3))^2-(sqrt(5))^2)`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = (1+sqrt(3)+sqrt(5))/(1+2sqrt(3)+3-5)`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = (1+sqrt(3)+sqrt(5))/(2sqrt(3)-1)`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = ((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/((2sqrt(3)-1)(2sqrt(3)+1))`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = ((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/((2sqrt(3))^2-1^2)`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = ((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/(12-1)`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = 1/11(1+sqrt(3)+sqrt(5))(2sqrt(3)+1)`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = 1/11(2sqrt(3)(1+sqrt(3)+sqrt(5))+1(1+sqrt(3)+sqrt(5)))`

`color(white)(1/(1+sqrt(3)-sqrt(5))) = 1/11(2sqrt(3)+6+2sqrt(15)+1+sqrt(3)+sqrt(5))`

`color(white)(1/(1+sqrt(3)-sqrt(5))) =(7+3sqrt(3)+sqrt(5)+2sqrt(15))/11`