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In the proof that [L^2,Lx] = 0 on page 161, how does Griffiths get from the first line to the second line?

He uses the commutation relation [AB,C] = A[B,C] + [A,C]B. The first term [Lx^2,Lx], is zero since Lx commutes with itself. Then using the commutation relation I introduced you expands [Ly^2,Lx] and [Lz^2,Lx]. Lets just expand [Ly^2,Lx], just to illustrate how the commutation relation works. In this case, A=Ly, B=Ly, and C=Lx. Using the relation we have [Ly*Ly,Lx] = Ly[Ly,Lx] + [Ly,Lx]Ly. He similar does this with [Lz^2,Lx].

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