pure math

f(x) = x^3 - 3bx + c
f'(x)=3x^2-3b
Consider f'(x)=0
x=√b
f(√b)=-2b√b+c
f(-√b)=2b√b+c
f(0)=c
f(+∞ )=+∞
f(-∞ )=-∞
The roots can only lies in the interval (i) +∞ to √b (ii) √b to -√b (iii) -√b to -∞
In order tp have three real roots, we need
f(√b)<0=>-2b√b+c<0=>c^2<4b^3=>4b^3 - c^2 > 0
f(-√b)>0=>2b√b+c>0=>4b^3 - c^2 > 0
Note: Since 2b√b is positive, from -2b√b<c<2b√b, we get |2b√b|>|c|, so we can take the square of both sides