2、(1)a(a-l) (a²+1)
(2) (2a-c+3b) (2a-c-3b)
(3) (a²+b²) (x²+y²)
3、解:(a+b+c)(a²+b²+c²-ab-bc-ca)
=(a³+ ab²+ac²-a²b-abc-a²c)+
(a²b+ b³+bc²-ab²-b²c- abc)+
(a²c+b²c+c- abc-bc²-ac²)
=a3³+b³ +c³- 3abc.
4、解:因为(n+7) ²- (n-5) ²
=[(n+7)+(n-5)][(n+7)-(n-5)]
= (n+7+n-5)(n+7-n+5)
=(2n+2)×12
=2×12(n+1)
=24(n+1),
且n为整数,所以(n+7) ²-(n-5) ²能被24整除.