$$n=3\cdot a-b^3$$ $$n=3a-b^3$$ $$is\ \ n^2+3\ divisible\ by\ 2\ ?$$ $$is\ \ n^2+3\ =even\ number\ $$
Since 3 is odd $$n^2\ has\ to\ be\ even\ number\ $$
Statement 1
$$a^2-4\cdot b^3-5=0$$ $$a^2-4b^3-5=0$$
$$a^2=0+4b^3+5$$ $$a^2=0+4b^3+5= even + even + odd=odd$$ $$4b^3=a^2-0-5=odd-even-odd=even$$ $$4b^3=even$$
But this does not tell us whether b is even or not
so, the statement is NOT SUFFICIENT.
Statement 2
$$3b^3-a^2+6=0$$ $$3b^3-a^2=-6$$
Neglecting the component
$$3b-a=even$$
This means both 3b and a are both odd. hence, it is either for b and a to be odd or both even
Considering $$n=3-b^3$$
If a and b are both even
So, $$n^3+3\ is\ not\ divisible\ by\ 2\ $$
Statement 2 alone is SUFFICIENT. $$answer\ is\ OPTION\ B$$
