Solver Equation If $A : D(A) \subset X \rightarrow X$ is a closed operator, then
\begin{eqnarray*}
R(\lambda : A) - R(\mu : A) = (\mu - \lambda) R(\lambda : A)R(\mu : A),  \ \ \forall \mu , \lambda \in \rho(A).
\end{eqnarray*}
 A: Start with $(I) = R(\lambda : A) - R(\mu : A) =  (\lambda I - A)^{-1} -  (\mu I - A)^{-1}$
You want to show $(I) = (\mu - \lambda)(\lambda I - A)^{-1}(\mu I - A)^{-1}$.
So multiply $(I)$ by $(\mu I - A)(\lambda I - A)$ and show the answer is $(\mu - \lambda)$.
A: For $\lambda \in \rho(A)$,
\begin{eqnarray}
AR(\lambda : A) & = & A(\lambda I - A)^{-1} \nonumber \\
& = & (A - \lambda I + \lambda I )(\lambda I - A)^{-1} \nonumber \\
& = & \lambda (\lambda I - A)^{-1} - (\lambda I - A)(\lambda I - A)^{-1} \nonumber \\
& = & \lambda (\lambda I - A)^{-1} - I \nonumber \\
& = & \lambda R(\lambda : A) - I.
\end{eqnarray}
From the above equality, it follows that for $\lambda , \mu \in \rho(A)$ the following identities are valid
\begin{eqnarray*}
R(\lambda : A) = R(\lambda : A)[ \mu R(\mu : A) - AR(\mu : A)]
\end{eqnarray*}
and
\begin{eqnarray*}
R(\mu : A) = R(\mu : A)[ \lambda R(\lambda : A) - AR(\lambda : A)]
\end{eqnarray*}
Subtracting the equations, we get
\begin{eqnarray*}
R(\lambda : A) - R(\lambda : A) & = & R(\lambda : A)[ \mu R(\mu : A) - AR(\mu : A)] \\
& - & R(\mu : A)[ \lambda R(\lambda : A) - AR(\lambda : A)] \\
& = & \mu R(\lambda : A)R(\mu : A) - AR(\lambda : A)R(\mu : A) \\
& - & \lambda R(\mu : A)R(\lambda : A) + AR(\lambda : A)R(\mu : A) \\
& = & (\mu - \lambda ) R(\lambda : A)R(\mu : A),
\end{eqnarray*}
