For all natural number n, 6n² + 6n is divisible by

​$$\text6n^2 + 6n$$

Factorization:
​$$\text a^2 + ab = a(a + b)$$​​
Divisibility rules:
A number is divisible by 6 if it is divisible by both 2 and 3

A number is divisible by 12 if it is divisible by both 4 and 3.

Solution:-
For  n=1:
​$$\text 6(1^2) + 6(1)$$ = 12 ( Divisible by both 6 and 12 but not by 18)

For n=2:

$$\text 6(2^2) + 6(2)$$= 36 ( Divisible by both 6 and 12 and also by 18)

For n =3:

​$$\text 6(3^2) + 6(3)$$ = 72 ( Divisible by both 6 and 12 and also by 18) 

As natural number = 1, 2, 3,........ 

So, ​$$\text6n^2 + 6n$$ will always be divisible by 6 and 12 

Hence, option (c) 6 and 12 both is the correct answer.