\frac{3^{a+3}-2 \cdot 3^{a+2}}{3^{a+1}+3^{a-1}}=
=\frac{3^a \cdot 3^3-2 \cdot 3^a \cdot 3^2}{3^a \cdot 3+3^a \cdot \frac{1}{3}}
=\frac{3^a \cdot 27 -3^a \cdot 18}{3^a(3+\frac{1}{3})}
=\frac{3^a(27-18)}{3^a \cdot 3\frac{1}{3}}
=\frac{9}{\frac{10}{3}}
=9 \cdot \frac{3}{10}
=\frac{27}{10} \in \mathbb W co kończy dowód