Răspuns :
[tex]\displaystyle\bf\\a)\\\\A = 3a^2 + 9c^2 + 6ab - 12ac - 6bc=\\\\=3(a^2 + 3c^2 + 2ab - 4ac - 2bc)= (Reordonam)\\\\=3(a^2 + 2 a b - 4 a c - 2 b c + 3 c^2)=(Repartizam: -4ac=-3ac-ac)\\\\=3(a^2 + 2 a b - 3 a c - ac - 2 b c + 3 c^2)=(Grupam)\\\\=3[(a^2 + 2 a b - 3 a c)+ (- ac - 2 b c + 3 c^2)]=(Dam~factor~comun)\\\\=3[a(a + 2b - 3c)-c(a + 2b - 3c)]=(Dam~factor~comun~paranteza)\\\\=\boxed{\bf3(a + 2b - 3c)(a-c)}[/tex]