= Solution
For $\lambda\ne\mu$, apply the <resolvent identity> from part (i) in both orders:
$$
R_\lambda R_\mu=\frac{R_\lambda-R_\mu}{\mu-\lambda},\qquad
R_\mu R_\lambda=\frac{R_\mu-R_\lambda}{\lambda-\mu}.
$$
The two quotients are equal. For $\lambda=\mu$ the assertion is immediate. Hence
$$
\boxed{R_\lambda R_\mu=R_\mu R_\lambda\qquad(\lambda,\mu>0).}
$$
Thus <resolvent operators> from the same <strongly continuous semigroup> commute. This conclusion does not require commuting arbitrary <bounded linear operators>; it uses the fact that both arise from the same <semigroup generator>.
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