$\underset{\mathrm{n}\text{ digits }}{(666\dots 6{)}^2+}\underset{\mathrm{n} \mathrm{digits}}{(888...8)}$  is equal to

$\begin{array}{l}(666\dots 6)=6+6\times 10+6\times {10}^2+\dots +6\times {10}^{\mathrm{n}-1}\\ \mathrm{n}\text{ digits }=6\left(1+10+{10}^2+\dots +{10}^{\mathrm{n}-1}\right)\\ =\frac{6}{9}\left({10}^{\mathrm{n}}-1\right)=\frac{2}{3}\left({10}^{\mathrm{n}}-1\right)\end{array}$

Similarly, $\underset{\mathrm{n}\text{ digits }}{(888\dots .})=\frac{8}{9}\left({10}^{\mathrm{n}}-1\right)$

Hence,  required sum $\begin{array}{l}=\frac{4}{9}{\left({10}^{\mathrm{n}}-1\right)}^2+\frac{8}{9}\left({10}^{\mathrm{n}}-1\right)\\ =\frac{4}{9}\left({10}^{2\mathrm{n}}-2\cdot {10}^{\mathrm{n}}+1+2\cdot {10}^{\mathrm{n}}-2\right)\\ =\frac{4}{9}\left({10}^{2\mathrm{n}}-1\right)\end{array}$