Factoring a difference of 2 cubes I am trying to factorize the expression $(a - 2)^3 - (a + 1)^3$ and obviously I would want to put it in the form of $(a - b)(a^2 + ab + b^2)$
So I start off with the first $(a - b)$ and I get $(a - 2) - (a + 1)$ which I simplify from $(a^2 + a -2a -2)$ to $(a^2 -3a -2)$
Now I'm up to $(a^2 + ab + b^2)$ and I $a^2$ would equal to $(a^2 - 4)$, $ab$ would be $(a - 2) * (a + 1)$ which is $(a^2 + a -2a -2)$  and $b^2$ would be $(a^2 + 1)$..
Then we get $(a^2 - 3a - 2)((a^2 - 4) (a^2 + a - 4a)(a^2 + 1)) $
At this point I get confused because I'm not sure if I did anything correct, and I don't know how to continue this. Any help is much appreciated
Regards,
 A: you have $$(a - 2)^3 - (a + 1)^3=[(a - 2) - (a + 1)][(a - 2)^2+(a - 2)(a + 1) + (a + 1)^2]
\\=[-3][a^2-4a +4+a^2 - a -2 + a^2 +2a+ 1]
\\=-3(3a^2-3a+3)
\\=-9(a^2-a+1)$$
A: Also in the second parenthesis $a$ squared will be $(a-2)$ squared i.e. $(a-2)(a-2) = a^2 -4a + 4$,
and similarly the $b$ squared will be $(a+1)$ squared i.e. $(a+1)(a+1)$ i.e. $a^2 + 2a + 1$.
I think having $a$ in the expression you want to factorize and in the identity you are trying to use is confusing you.  Why not instead factorize $(x-2)^3 - (x+1)^3$?
Then in your identity $a$ is $(x-2)$ and $b$ is $(x+1)$ and you get
$( (x-2)-(x+1) )( (x-2)^2 + (x-2)(x+1) + (x+1)^2)$
which finally simplifies down to $-3(3x^2 - 3x +3)$ which gives
$-9(1 - x +x^2)$.
Then turn it back to $a$ at the end to get
$-9(1 - a +a^2)$
and as a check, try $a = 0$: 
$(0-2)^3 -(0+1)^3 = -8 - 1 = -9$
$-9(1 - 0 + 0^2) = -9(1) = -9$ 
A: Brute force algebra gives us
$$
(a-2)^3 - (a+1)^3 
= (a^3 -6a^2 + 12a - 8) - (a^3 + 3a^2 + 3a +1)
= -9(a^2 - a + 1)
$$
But $a^2 - a + 1$ doesn't have any factors (not real ones, anyway).
