Metformin
(a2+2b2-2ab)(a2+2b2+2ab)
24a + b2 + 3a + 2b2= 27a + 3b2
do you mean 12a3 - 5a2b - 2b2 ?? = 12a3 + 8a2b - 3a2b - 2b2 then group 1st 2 terms and last 2 and use GCF = (12a3 + 8a2b) + ( - 3a2b - 2b2) = 4a2(3a + 2b) + (-b)(3a2 +2b) not much help, but that's all folks
6b2c3
We can combine the like terms. So the b2 can be combined with the 2b2 to give 3b2. Likewise the 3b plus the -5b gives -2b.Therefore, b2 + 3b - 5b + 2b2 = 3b2 - 2b.
2b2 + 8 para b = -3
n nn n
3a2b(2b2-1)
= 4a2 + 2ab 2b2
3(2b2 - 5b - 2)
11(2b2 + 21)
C. 12b3