I'm trying to fill in the gaps in my knowledge of simplifying rational expressions using conjugates, but this one stumps me. Given $\tan(\frac{\pi}{4}-\frac{\pi}{6})$, I can work the formula down to:
$$\frac{9-3\sqrt{3}}{9+3\sqrt{3}}$$
Then I attempt to multiply top and bottom by the denominator's conjugate $9-3\sqrt{3}$, only to get something like:
$$\frac{-27\sqrt{3}-27\sqrt{3}+108}{54}$$
which of course is wrong. The correct answer should be $2-\sqrt{3}$. Thanks in advance for any help. I simply can't figure out where my distribution is going wrong...
The arithmetic will be simpler if we immediately cancel a $3$. Even better is to cancel a $3\sqrt{3}$. If we do that, we get $$\frac{\sqrt{3}-1}{\sqrt{3}+1}.$$ Multiply top and bottom by $\sqrt{3}-1$. At the bottom we get $2$. At the top, we get $(\sqrt{3}-1)^2$, which is $3-2\sqrt{3}+1$, that is, $4-2\sqrt{3}$. finally, divide by $2$.
Remark: Your answer was not wrong. You got $\frac{108-54\sqrt{3}}{54}$. This is $\frac{108}{54}-\frac{54\sqrt{3}}{54}$, which is $2-\sqrt{3}$.
For the original problem, it would have been easier to use the formula $$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y},$$ a mild variant of the more familiar $\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$.