alternative expression of L^2

show that $\displaystyle L^2$ may be alternative expressed as

$\displaystyle L^2 = L_{0}^{2} + L_0 + L_{-}L_{+} = L_{0}^{2} - L_0 + L_{+}L_{-}$

and that $\displaystyle [L^2, L_{\alpha}] = 0$ for $\displaystyle \alpha = 1, 2, 3$

i know that the generators are $\displaystyle L_{\alpha} = ix_{\alpha}$

and $\displaystyle L_+ = L_1 + iL_2, L_- = L_1 - iL_2$

but i can't get $\displaystyle L^2 = L_{0}^{2} = L_0 + L_{-}L_{+} = L_{0}^{2} - L_0 + L_{+}L_{-}$ no matter which way i try it!

what is $\displaystyle L^2$ anyway? is it $\displaystyle (x_1 x_2 x_3)^2$?

Re: alternative expression of L^2

or is $\displaystyle L^2 = L_1^2 + L_2^2 + L_3^2$?