From Förberedande kurs i matematik 1

Because \displaystyle 5^{9}=5^{8+1}=5^{8}\centerdot 5^{1}=5^{8}\centerdot 5, the two terms inside the brackets have \displaystyle 5^{8} as a common factor and can therefore be taken outside the bracket.

\displaystyle \begin{align} & \left( 5^{8}+5^{9} \right)^{-1}=\left( 5^{8}+5^{8}\centerdot 5 \right)^{-1}=\left( 5^{8}\centerdot \left( 1+5 \right) \right)^{-1} \\ & \\ & =\left( 5^{8}\centerdot 6 \right)^{-1}=5^{8\centerdot \left( -1 \right)}\centerdot 6^{-1}=5^{-8}\centerdot 6^{-1}. \\ \end{align}

Furthermore, \displaystyle 625=5\centerdot 125=5\centerdot 5\centerdot 25=5\centerdot 5\centerdot 5\centerdot 5=5^{4} and we obtain

\displaystyle \begin{align} & 625\centerdot \left( 5^{8}+5^{9} \right)^{-1}=5^{4}\centerdot 5^{-8}\centerdot 6^{-1}=5^{4-8}\centerdot 6^{-1} \\ & \\ & =5^{-4}\centerdot 6^{-1}=\frac{1}{5^{4}}\centerdot \frac{1}{6}=\frac{1}{5^{4}\centerdot 6}=\frac{1}{5\centerdot 5\centerdot 5\centerdot 5\centerdot 6} \\ & \\ & =\frac{1}{3750} \\ \end{align}