Question commutation in quantum mechanics

[Another]

Homework Statement

Show that $[L_{x}^2,L_{y}^2]=[L_{y}^2,L_{z}^2]=[L_{z}^2,L_{x}^2]$

Homework Equations

$L^2 = L_{x}^2+L_{y}^2+L_{z}^2$
$L_x = yp_z-zp_y$
$L_y = zp_x-xp_z$
$L_z = xp_y-yp_x$
$[x_i,p_j]=iħδ_{ij}$
$[L_x,L_y]=iħL_z$
$[L_y,L_z]=iħL_x$
$[L_z,L_x]=iħL_y$
$[A,B]=AB-BA$

The Attempt at a Solution

$[L_{x}^2,L_{y}^2]= [L_{x}L_{x},L_{y}L_{y}]$
$[L_{x}^2,L_{y}^2]= L_{x}L_{x}L_{y}L_{y}-L_{y}L_{y}L_{x}L_{x}$
$[L_{x}^2,L_{y}^2]= L_{x}L_{x}L_{y}L_{y}-L_{y}L_{y}L_{x}L_{x}±L_{y}L_{x}L_{y}L_{x}$
$[L_{x}^2,L_{y}^2]= L_{x}L_{x}L_{y}L_{y}-L_{y}L_{x}L_{y}L_{x}+L_{y}L_{x}L_{y}L_{x}-L_{y}L_{y}L_{x}L_{x}$
$[L_{x}^2,L_{y}^2]= L_{x}L_{x}L_{y}L_{y}-L_{y}L_{x}L_{y}L_{x}+L_{y}(L_{x}L_{y}-L_{y}L_{x})L_{x}$
$[L_{x}^2,L_{y}^2]= L_{x}L_{x}L_{y}L_{y}-L_{y}L_{x}L_{y}L_{x}+L_{y}[L_{x},L_{y}]L_{x}$
$[L_{x}^2,L_{y}^2]= L_{x}L_{x}L_{y}L_{y}-L_{y}L_{x}L_{y}L_{x}+iħL_{y}L_{z}L_{x}$

$[L_{y}^2,L_{z}^2]= [L_{y}L_{y},L_{z}L_{z}]$
$[L_{y}^2,L_{z}^2]= L_{y}L_{y}L_{z}L_{z}-L_{z}L_{z}L_{y}L_{y}$
$[L_{y}^2,L_{z}^2]= L_{y}L_{y}L_{z}L_{z}±L_{z}L_{y}L_{z}L_{y}-L_{z}L_{z}L_{y}L_{y}$
$[L_{y}^2,L_{z}^2]= L_{y}L_{y}L_{z}L_{z}-L_{z}L_{y}L_{z}L_{y}+L_{z}L_{y}L_{z}L_{y}-L_{z}L_{z}L_{y}L_{y}$
$[L_{y}^2,L_{z}^2]= L_{y}L_{y}L_{z}L_{z}-L_{z}L_{y}L_{z}L_{y}+iħL_{z}L_{x}L_{y}$

$[L_{z}^2,L_{x}^2]= [L_{z}L_{z},L_{x}L_{x}]$
$[L_{z}^2,L_{x}^2]= L_{z}L_{z}L_{x}L_{x}-L_{x}L_{x}L_{z}L_{z}$
$[L_{z}^2,L_{x}^2]= L_{z}L_{z}L_{x}L_{x}±L_{x}L_{z}L_{x}L_{z}-L_{x}L_{x}L_{z}L_{z}$
$[L_{z}^2,L_{x}^2]= L_{z}L_{z}L_{x}L_{x}-L_{x}L_{z}L_{x}L_{z}+L_{x}L_{z}L_{x}L_{z}-L_{x}L_{x}L_{z}L_{z}$
$[L_{z}^2,L_{x}^2]= L_{z}L_{z}L_{x}L_{x}-L_{x}L_{z}L_{x}L_{z}+iħL_{x}L_{y}L_{z}$

It not equal. My answer is incorrect.

 

[jshtok]

I suggest to use the commutator identities:
[AB,C] = A[B,C] + [A,C]B
[A,BC]=B[A,C] + [A,B]C
(one possible source: http://www.cchem.berkeley.edu/chem120a/extra/commutator.pdf)

Thus, [Lx²,Ly²] = [LxLx, Ly²] = Lx[Lx,Ly²]+[Lx,Ly²]Lx
Then you compute the [Lx,Ly²], which decomposes as a sum of products of Ly and [Lx,Ly].
Notice that the latter do not commute, as [Lx,Ly] ~ Lx, but the whole thing unravels.

I didn't try to go to the end with this, but I believe it should work.

Regards,
Joseph Shtok