Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there a formula for expanding these kind of things (what kind of expression is it, by the way?)?

This is my solution:
$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$
$= \displaystyle\frac{[(\sqrt5+2)^3+(\sqrt5-2)^3][(\sqrt5+2)^3-(\sqrt5-2)^3]}{8\sqrt5}$
$=\displaystyle\frac{(\sqrt5+2+\sqrt5-2)[(\sqrt5+2)^2\color{red}{+}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2](\sqrt5+2-\sqrt5+2)[(\sqrt5+2)^2\color{red}{-}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2]}{8\sqrt5}$
$=\displaystyle\frac{[2\sqrt5(5+4\sqrt5+4+\color{red}{5-4}+5-4\sqrt5+4][4(5+4\sqrt5+4\color{red}{-(5-4)}+(5-4\sqrt5+4)]}{8\sqrt5}$
$=\displaystyle\frac{2584\sqrt5}{8\sqrt5}$
$=323$
Because of the multiplication, I still got the same answer as given in the book. However, is the book or I correct in terms of the positive/negative signs(in red)?

It might be interesting that you may avoid tedious calculations with roots if you use recurrence relations:


*

*Set $t_1 =2+\sqrt 5$ and $t_2 = 2-\sqrt 5$.


So, the searched for value is 
$$\frac{t_1^6-t_2^6}{8\sqrt{5}}$$
This is $a_6$ in the recurrence relation 
$$a_{n+2} - (t_1+t_2)a_{n+1} + t_1t_2 = a_{n+2} - 4a_{n+1} -a_n $$
with 
$$a_0 = 0 \text{ and } a_1 = \frac{t_1-t_2}{8\sqrt{5}}=\frac 14$$
Now, calculating recursively you get 
$$a_6 = 4\left(76+\frac 14\right)+18 = 323$$
