Cubic polynomial discriminant I am stuck on a cryptography problem that pertains to Elliptic Curves.
The problem is stated as follows:
Assume the cubic polynomial $X^3+AX+B = (X-a)(X-b)(X-c)$
If $4A^3 + 27B^2 = 0$, then show two or all of $a,b,c$ are the same. 
So far, I expanded the right side, so I get the following equations:
$0 = a + b + c$ 
$A = ab + ac + bc$ 
$B = -abc$
I can't seem to use the hypothesis in the correct way. I tried to compute $A^3$ but the expansion of it looks horrible. If any one has any tips how to approach this problem, that would be great. 
 A: The case $A=0$ must be treated separately: In this case, $X^3+B$ has a double root if and only if $27B^2=0$, i.e. if and only if $B=0$. That is obvious, so let us assume $A\ne 0$.
Lets set $f(X):=X^3+AX+B$ and consider $f'(X)=3X^2 + A$. A polynomial has a double root if and only if it shares this root with its derivative. Hence, let us check when that is the case. $f'(x)=0$ means $x^2 = -\frac 13 A$ and $f(x)=0$ means $0=x^3+Ax+B=-\frac{1}{3}Ax+Ax+B$, so $\frac{2}{3}Ax = -B$, i.e. $x=\frac{-3B}{2A}$. Plugging this back into $f'$, we get that 
$$0=f'(x)=3\cdot\frac{9B^2}{4A^2} + A,$$
which yields $27B^2 = -4A^3$. Hence, $f$ has a double root if and only if $27B^2+4A^3=0$.
A: Let $X^3+Ax+B=(X-p)(X-q)^2$
Comparing the coefficients of the different powers of X,
$p+2q=0=>p=-2q$
$2pq+q^2=A=>A=-3q^2$
$-pq^2=B=>B=-q^2(-2q)=2q^3$
Eliminating q, $4A^3+27B^2=0$  as $A^3=-27q^6$  and $B^2=4q^6$

Conversely, the parametric values of (A,B) can be written as $(-3s^2,2s^3)$
Then, the equation becomes $X^3-3s^2X+2s^3=0$
Clearly, s is one of the solutions.
On the division by (X-s), we get $X^2+sX-2s^2=0 =>X=s,-2s$
