= Solution
The <root-space decomposition> gives $[L_r,L_s]\subseteq L_{r+s}$ when $r+s\ne0$. The two <roots> are independent, so $r+s\ne0$, and the missing root sum means that this weight space is zero. Hence $[e_r,e_s]=0$. These adjoint root operators are nilpotent: successive brackets shift root weights by the same root, so finite <root strings> force eventual vanishing; on the <Cartan subalgebra> the first bracket lies in its root space and the next is zero. By part (iv), the <nilpotent operators> $A=t\operatorname{ad}e_r$ and $B=u\operatorname{ad}e_s$ commute. Every power of $A$ therefore commutes with every power of $B$, so their finite <matrix exponentials> commute too. Using the inverses proved in part (ii),
$$
\boxed{x_s(u)^{-1}x_r(t)^{-1}x_s(u)x_r(t)=I.}
$$
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