Simply laced Chevalley commutator formula ($x_s(u{)}^{-1}{x}_r(t{)}^{-1}{x}_s(u)x_r(t)=x_{r+s}(-N_{r,s}tu)$)

For equal-length roots $r,s$ with a root sum, their angle is $2\pi /3$. The vectors $2r+s$ and $r+2s$ have squared length three times a root's squared length, so neither is a root. Thus $[e_r,e_{r+s}]=[e_s,e_{r+s}]=0$. Apply the central-commutator exponential identity to $A=t\mathrm{ad}{e}_r$ and $B=u\mathrm{ad}{e}_s$, whose commutator is $tu{N}_{r,s}\mathrm{ad}{e}_{r+s}$, to obtain the displayed identity over the complex numbers.