= Simply laced Chevalley commutator formula
{title2=$x_s(u)^{-1}x_r(t)^{-1}x_s(u)x_r(t)=x_{r+s}(-N_{r,s}tu)$}
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 $A=t\operatorname{ad}e_r$ and $B=u\operatorname{ad}e_s$, whose commutator is $tuN_{r,s}\operatorname{ad}e_{r+s}$, to obtain the displayed identity over the complex numbers.
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