It is known that a + B = 2008, C = 2007 Find: 2A (a + B-C) - 2b (C-A-B) - C (B-C + a)

It is known that a + B = 2008, C = 2007 Find: 2A (a + B-C) - 2b (C-A-B) - C (B-C + a)

2a(a+b-c)-2b(c-a-b)-c(b-c+a)
=2a(a+b-c)+2b(a+b-c)-c(a+b-c)
=(a+b-c)(2a+2b-c)
=(a+b-c)[a+b-c+(a+b)]
=(2008-2007)(2008-2007+2008)
=2009

(a ^ 2 + 9) ^ 2-36a ^ 2 extract common factor

Original formula = (a ^ 2 + 9) ^ 2 - (6a) ^ 2
=(a^2+9+6a)(a^2+9-6a)
=(a+3)^2(a-3)^2

13 (2) the problem of extracting common factor
Fill in the blanks:
Square of 2m (Y-X) = () (X-Y)
-2m (Y-X) cube = () (X-Y) cube
（x+y)(y-x)=-(x+y)( )
-M (Y-X) = m ()
Factorization:
3a(m-n)-2b(m-n)
a(x-2)-b(x-2)+(2-x)
The square of 9 (a-b) (a + b) - 3 (a-b)
-Square of 2 (b-2a) + (b-2a)
The square of 3 (x + y) (Y-X) - (X-Y)
-3 (X-Y) square - (Y-X) cube

Is it all right? Is there any decomposition factor that can give process?