Second-order lie bracket can be expressed as a linear combination of vector fields.

Question

I am new to lie brackets and I am stuck on how to formulate this problem. Given: $$\dot x=f(x)+ug(x),\quad x \in \Omega \subset \Bbb R^3, \quad |u| \le 1, \quad f,g \in C^\infty \\$$ Where $\Omega$ is a domain (open and connected set) on which the vector fields $f,g$ and their Lie brackets $[f,g]$ are linearly independent. Let $X=f-g$ and $Y=f+g$ denote the vector fields corresponding to the constant controls $u\equiv -1$ and $u\equiv 1$, respectively.$$ $$ Show that on $\Omega$ the second-order Lie brackets $[X,[f,g]]$ and $[Y,[f,g]]$ can be expressed as linear combinations of the vector fields $f,g$ and $[f,g]$, $$[X,[f,g]]=\alpha f+a_1 g+a_2[f,g],\quad [Y,[f,g]]=\beta f+b_1 g+b_2 [f,g]$$ with coefficeints $\alpha,\beta$ and $a_i,b_i, i=1,2$ which are $C^\infty$ functions on $\Omega$.$$ $$ This is what I have done so far: $$[f,g] = L_f\circ L_g-Lg\circ L_f=Z$$ $$[X,Z] = L_X \circ L_Z-L_Z \circ L_X = L_X\circ L_f\circ L_g-L_X\circ L_g\circ L_f-L_f\circ L_g\circ L_X+L_g\circ L_f\circ L_X$$ This then can be split out into $[f,[f,g]]-[g,[f,g]]$: $$\Rightarrow L_f\circ L_f\circ L_g-L_f\circ L_g\circ L_f-L_f\circ L_g\circ L_f+L_g\circ L_f\circ L_f-(L_g\circ L_f\circ L_g-L_g\circ L_g\circ L_f-L_f\circ L_g\circ L_g+L_g\circ L_f\circ L_g)$$ $$\Rightarrow L_f\circ L_f\circ L_g-2L_f\circ L_g\circ L_f+L_g\circ L_f\circ L_f\\-(2L_g\circ L_f\circ L_g-L_g\circ L_g\circ L_f-L_f\circ L_g\circ L_g) $$ This is as far as I have gotten. I am not sure if I am going down the right path or not. Any help would be appreciated.