Let g be a Lie algebra such that [[x,y],y]=0 for all $x,y \in g$. Show that 3[[x,y],z]=0 for all $x,y,z \in g$. [Hint: Observe that the mapping (x,y,z) to [[x,y],z] is skew-symmetric in x,y,z and make use of the jacobi identity.]
So I know that [[x,y],z]=[x,[y,z]]+[y,[z,x]] (using the jacobi identity)
So 3[[x,y],z]=3[x,[y,z]]+3[y,[z,x]], I need to use [[x,y],y]=0, but can't see how to do it. Like I would need to elimate x,y or z from 3[x,[y,z]]+3[y,[z,x]], but it just seems impossible.
Was wondering am I using skew symmetric wrong. I assume that skew symmetric is just this [x,y]=-[y,x].