Your notation is confusing. It looks like: 8a3 + 4a2b - 2ab2 - b3. So if you look at it, notice 23 = 8 & 22 = 4, so take c = (2*a), and you have c3 + c2b - cb2 - b3, Now note that if c = b, the polynomial is zero, so (c-b) or (b-c) is a factor. So divide by (c-b) using long division: (c - b)(c2 + 2cb + b2) {the second polynomial is a perfect square: (c+b)2}, so (c-b)(c+b)2, then substitute c = 2a, and get (2a-b)(2a+b)2 finally, multiply it out to make sure you get the original polynomial.
(2a + b)(2a + b)(2a - b)
4(a + 2b - 4c)
4a + 3c - 2b - c + a - b = 5a - 3b + 2c
a-b+a-b+2b+a = 4a
a+a+a+a+b+b - add the like terms =4a+2b 4a and 2b cannot be added because they are not like terms.
4a+2b-1a+9b combine all alike variables using commutative property of addition 4a-1a+2b+9b Then add the like variables 3a+11b
4a+2b+a
3a - 2b
7a-2b
10a - 10 b
= 4a2 + 2ab 2b2
It is 2a-b simplified
Assuming the missing sign is "minus" then you have the difference between two squares. This is solved as (x + y)(x - y). In your example x = 4a and y = 2b so the factorisation is (4a + 2b)(4a - 2b) or more completely: (4)(2a + b)(2a - b).