Consider the commutative ring Z36, under + and multi mod 36. For any ideal A of Z36, we define the nil radical of A as:
nil(A) = {r belong to Z36 : r^n belong to A for some positive integer n}. Note, n is not fixed. It depends on the element r belong to Z36.

(a) Choose an element x belong to Z36 and determine the principal ideal (PLEASE solve this for the element x=3)
(b) For the principal ideal you generated in (a), determine nil().
(c) Let A be any ideal of Z36. Prove that nil(A) is an ideal of Z36

(PLEASE solve this for the element x=3)