Find condition for $ a $ so that $ x^3 - 3ax + 1$ is separable Let $ K $ be an algebraically closed field of characteristic $ 0 $. What is a polynomial condition for $ a \in K $ such that $ f(x) = x^3 - 3ax + 1$ has distinct roots?
Here is my attempt: note that $ Df(x) = 3x^2 - 3a $. If $ a = 0 $, then $ (x^3 + 1,  3x^2 ) = 1 $, hence $ f $ is separable. Otherwise, it suffices to consider $ (x^3 - 3ax +1,  x^2 - a) $. Using polynomial division, we have $$ (x^3 - 3ax +1,  x^2 - a) = (x^2 - a, 2ax - 1) = (x-1/(2a), a - 1/(4a^2)). $$
Hence $ f $ is separable if $ a - 1/(4a^2) \neq 0 $, i.e. $ 4a^3 \neq 1 $. So the polynomial condition for $ a $ is that $ a $ does not satisfy $ 4x^3 - 1 = 0 $. Is this correct?

Your computation of the greatest common divisor is wrong.
The division of $x^3-3ax+1$ by $x^2-a$ has remainder $-2ax+1$.
Now we can divide $4a(x^2-a)$ by $2ax-1$, which has remainder $1-4a^3$.
(The case $a=0$ is trivial, so multiplying by a constant doesn't change the greatest common divisor, which is determined up to a multiplicative constant.)
