In general, P(X) is separable if and only if it is square-free over any field that contains K, which holds if and only if P(X) is coprime to its formal derivative D P(X).
This is actually a typical example of why inseparability matters; in geometric terms P represents the mapping on the projective line over the finite field, taking co-ordinates to their pth power.
It is of degree p, but has no automorphism fixing K, other than the identity, because T 1/p is the unique root of P. This shows directly that Galois theory must here break down.
One can show that the tensor product of fields of L with itself over K for this example has nilpotent elements that are non-zero.
If P(x) is separable, and its roots form a group (a subgroup of the field K), then P(x) is an additive polynomial.
For example, let P be an irreducible polynomial with integer coefficients and p be a prime number which does not divide the leading coefficient of P. Let Q be the polynomial over the finite field with p elements, which is obtained by reducing modulo p the coefficients of P. Then, if Q is separable (which is the case for every p but a finite number) then the degrees of the irreducible factors of Q are the lengths of the cycles of some permutation of the Galois group of P. Another example: P being as above, a resolvent R for a group G is a polynomial whose coefficients are polynomials in the coefficients of P, which provides some information on the Galois group of P. More precisely, if R is separable and has a rational root then the Galois group of P is contained in G. For example, if D is the discriminant of P then