I'm trying to factorise x^6+x^3+1 into irreducible factors in the finite field F19 (integers mod 19)- it seems time consuming and complicated to check each member of F19 to see if it's a root. Is there an easier way to do it?

Well, for this particular problem, you could let y= x^3 so that your polynomial becomes y^2+ y+ 1. And now note that 1- 3(19)= -56= 1 (mod 19) so that this is the same as y^2+y-56= (y+8)(y- 7).