TOC closure property doubt closed

Question

If a language L1 is given as aⁿbⁿ  and L2 is given as {a,b}* , then the language L1 - L2 will be : regular or CFL and why ?
My doubt is that since L2 is a regular language and L1 is CFL and L2 will contain all strings in L1, so L1 - L2 will give {} which is regular. 
But I found in some book the since L1 is CFL and L2 is regular L1 - L2 can be written as $L1\cap L2¯$ ( CFL intersection regular complement ). Complement of regular is regular and intersection of CFL with regular is closed and the language will be CFL.

Which one is right and why ?

Comments

Both are right.

If a language is regular then it is also CFL. Since regular is a subset of CFL.

So, L1-L2 is regular and CFL also.

---

Ok, got it thanks.

---

here L1 is CFL and L2 is Regular(CFL also)...and

L1-L2(CFL-CFL) difference of two CFL are not closed(it means that language obtain from 'CFL-CFL'  is not CFL for always but for some language it may be..

for given language L1-L2 returns phi  (which is regular )....

Answer 1

Remember that all languages are closed in following operations:

1)Intersection with RL (L union R)

2)Difference with RL (L - R)