Also...
\(\frac{(-3)^{2n+1}}{27\,\cdot\,(-3)^{2n}}=\frac{(-3)^{2n+1}}{27}\,\cdot\,\frac{(-3)^{2n+1}}{(-3)^{2n}}\)
This is not true.
If you multiply the two fractions on the right side together, you will get \(\frac{[ (-3)^{2n+1})]^2}{27\,\cdot\,(-3)^{2n}}\) .
\(\frac{a}{bc}\,\neq\,\frac{a}{b}\,\cdot\,\frac{a}{c}\)
But..you can distribute division the same as you distribute multiplication, like this....
\(\frac{8 + 6 +10}{2}=\frac12(8+6+10)\,=\,(\frac12)(8)+(\frac12)(6)+(\frac12)(10)\,=\,4+3+5\,=\,12\) 