a, (n²+ 3n-1)(n+2)- n³+2
= n³+ 2n²+ 3n²+ 6n-n- 2- n³+2
= 5n²+ 5n
= 5n(n+1) chia hết cho 5
=> (n²+ 3n-1)(n+2)- n³+2 chia hết cho 5
b, (6n+1)(n+5)- (3n+5)(2n-1)
= 6n²+ 20n+ n+ 5 - (6n²- 3n+ 10n- 5)
= 6n²+ 21n+ 5- 6n²+ 3n- 10n+5
= 14n+ 10
= 2( 7n+5) chia hết cho 2
=> (6n+1)(n+5)- (3n+5)(2n-1) chia hết cho 2