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)?
 A: The book is correct. Notice the signs in the identities:
$$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$
$$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$
Let $a = (\sqrt{5}+2)^2$ and $b = (\sqrt{5}-2)^2$ and plug in to the second formula to recover your equation.
Your arithmetic happened to work by the lucky circumstance of $(18+1)(18-1)$ equalling $(18-1)(18+1)$ 
A: The book solution used the formulas for the sum and difference of two cubes, 
$x^3+y^3=(x+y)(x^2-xy+y^2)$ and $x^3-y^3=(x-y)(x^2+xy+y^2),$
with $x=\sqrt5+2$ and $y=\sqrt5-2$.
A: Hint
$a-b=4$
$a+b=2\sqrt5$
$ab=1$
$a^3-b^3=(a-b)^3+3ab(a-b)=?$
$a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a-b)^2+ab)=?$
A: 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 
$$a_6 = 4\left(76+\frac 14\right)+18 = 323$$
A: Alternatively, let $a=9+4\sqrt5$ and $b=9-4\sqrt5$. Then, $a+b=18$, $ab=1$ and,
$$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}
=\frac{a^3 - b^3}{8\sqrt5}=\frac{(a-b)(a^2 +ab+b^2)}{8\sqrt5}$$
$$=\frac{\sqrt{(a+b)^2-4ab}\>[(a+b)^2 -ab)]}{8\sqrt5}
=\frac{\sqrt{18^2-4}\>(18^2 -1)}{8\sqrt5}=323$$
