okie, let's see:
from 2, we obtain b=4 or -1
that is to say : the square root can be 7^2+(-1) (= 48) or 7^2+ 2(=51)
we have a^3-a^2 = 48
---> a^3 + 3a^2- 4a^2 + 12a -12a -48= 0 ( this is a high school's method to factorize the expression to obtain an equation with RHS= 0)
---> a^2(a-4) + 3a(a-4) +12(a-4)= 0
----> (a-4) * ( a^2+3a+12)=0
-----> a=4, a^2 + 3a +12 has no solution since delta<0 ( quadratic equation method)
case 2 : a^3- a^2= 51 or a^3-a^2-51=0
since a must be an integer ---> the possible roots of this equation must be among factors of 51 ( +-17 and +- 3) . You can plug these numbers to see if they're roots of the equation
. NOTE: Better method to plug is to right away notice that the root can't be negative because the highest power of a here, a^3 determine the value of LHS. After plugging, you find that there's no root for this equation.
Thus, the only value of a obtained is 4 when b=-1
Thus, B is correct!