Simplifying $\frac { \sqrt2 + \sqrt 6}{\sqrt2 + \sqrt3}$?

Question

Is there anything else you can do to reduce it to something "nicer" other than multiplying it by $\dfrac {\sqrt3 - \sqrt2}{\sqrt3 - \sqrt2}$ and get $\sqrt 6 -2 + \sqrt {18} - \sqrt {12}$? The reason I think there's a nicer form is because the previous problem in the book was to simplify $\sqrt{3+ 2 \sqrt 2} - \sqrt{3 - 2 \sqrt 2}$, which simplifies nicely to $\sqrt{( \sqrt 2+1)^2} - \sqrt{ ( \sqrt 2 - 1)^2} = 2.$

Answer 1

Well, you could note that

$$\sqrt6-2+\sqrt{18}-\sqrt{12}=-2+\sqrt6(1-\sqrt2+\sqrt3)$$

But beyond that, it doesn't look any better.

Answer 2

You are correct. Multiply by the conjugate which is $\sqrt{2}-\sqrt{3}$

After multiplying by the conjugate we have $\frac{2-\sqrt6+\sqrt{12}-\sqrt{18}}{-1}$ which gives $-2+\sqrt6-\sqrt{12}+\sqrt{18}$. The only further simplification is as follows: Simplify $\sqrt{12}$ and $\sqrt{18}$ and we have $-2+\sqrt6-2\sqrt3+3\sqrt2$